The Hardy Inequality for Hermite Expansions

Abstract

The purpose of the paper is to prove a sharp form of Hardy-type inequality, conjectured by Kanjin, for Hermite expansions of functions in the Hardy space $H^1({\mathbb {R}})$, that is, $\sum _{n=1}^{\infty }n^{-\frac{3}{4}}|a_n(f)|\le A\Vert f\Vert _{H^1({\mathbb {R}})}$ for all $f\in H^1({\mathbb {R}})$, where $A$ is a constant independent of $f$.

1 Introduction and Result

The Hermite polynomials $H_n(x)$ ($n\ge 0$) are defined by the orthogonal relation (cf. [15, 16])

$$\begin{aligned} \int _{-\infty }^{\infty }e^{-x^2}H_n(x)H_m(x)dx=\pi ^{\frac{1}{2}}2^n n!\delta _{nm}, \end{aligned}$$

and the Hermite functions are given by

$$\begin{aligned} {\mathcal {H}}_n(x)=\left( \pi ^{\frac{1}{2}}2^n n!\right) ^{-\frac{1}{2}}e^{-\frac{x^2}{2}}H_n(x), \quad n=0,1,2,\ldots , \end{aligned}$$

which are orthonormal over $(-\infty ,\infty )$ associated with the Lebesgue measure.

For $f\in L({\mathbb {R}})$, its Hermite expansion is

$$\begin{aligned} f(x)\sim \sum _{n=0}^{\infty }a_n(f){\mathcal {H}}_n(x),\quad a_n(f)=\int _{-\infty }^{\infty }f(x){\mathcal {H}}_n(x)dx. \end{aligned}$$

The Hardy type inequality for Hermite expansions of functions in the Hardy space $H^1({\mathbb {R}})$ has been studied in several works. The first step was done by Kanjin [5] who proved the inequality

$$\begin{aligned} \sum _{n=1}^{\infty }n^{-\frac{29}{36}}|a_n(f)|\le A\Vert f\Vert _{H^1({\mathbb {R}})}, \quad f\in H^1({\mathbb {R}}), \end{aligned}$$

(1)

where $A>0$ is a constant independent of $f$. Balasubramanian and Radha [2] extended Kanjin’s result to $H^p({\mathbb {R}})$, $0<p\le 1$. Radha and Thangavelu [10] (cf [17] also) obtained inequalities of Hardy type for $d$-dimensional Hermite and special Hermite expansions for $d\ge 2$, where the constant they determined, in place of $\frac{29}{36}$, is

$$\begin{aligned} \sigma =\left( \frac{d}{2}+1\right) \left( \frac{2-p}{2}\right) \end{aligned}$$

for $H^p({\mathbb {R}}^d)$, $0<p\le 1$. In comparison with the case of $d$-dimension ($d\ge 2$), the Hardy inequality for one-dimensional Hermite expansions should be the one as (1) but with $\frac{3}{4}$ instead of $\frac{29}{36}$. However the method in [10] does not work for $d=1$. An improved form of (1) with $\frac{3}{4}+\epsilon$ for $\epsilon >0$ in place of $\frac{29}{36}$ was obtained by Kanjin [6]. Moreover Kanjin [6] proved his inequality for all $f\in L^1({\mathbb {R}})$, that is,

$$\begin{aligned} \sum _{n=1}^{\infty }n^{-\frac{3}{4}-\epsilon }|a_n(f)|\le A\Vert f\Vert _{L^1({\mathbb {R}})}. \end{aligned}$$

(2)

This again leads Kanjin to conjecture that the possible form of the Hardy inequality for Hermite expansions would be

$$\begin{aligned} \sum _{n=1}^{\infty }n^{-\frac{3}{4}}|a_n(f)|\le A\Vert f\Vert _{H^1({\mathbb {R}})}. \end{aligned}$$

(3)

We shall give a positive answer to this conjecture in the present paper. Kanjin [6] also showed that there exists a function $f_0\in L^1({\mathbb {R}})$ such that

$$\begin{aligned} \sum _{n=1}^{\infty }n^{-\frac{3}{4}}|a_n(f_0)|=\infty , \end{aligned}$$

so the Hardy space norm is required in (3).

The proofs of (1) and (2) in [5, 6] were based on the pointwise estimate of the Hermite functions as follows: for given $\tau >0$, there exist positive constants $A$, $\eta$ and $\xi$ such that

$$\begin{aligned} |{\mathcal {H}}_n(x)|\le A\left( |x|+\sqrt{n}\right) ^{-\frac{1}{4}} \left( ||x|-\sqrt{2n}|+n^{-\frac{1}{6}}\right) ^{-\frac{1}{4}} \Psi _n(x) \end{aligned}$$

(4)

holds for all $x\in {\mathbb {R}}$ and $n\ge 1$, where

$$\begin{aligned} \Psi _n(x)=\left\{ \begin{array}{ll} 1,&{}\quad \hbox {for}\quad 0\le |x|\le \sqrt{2n};\\ \exp \left( -\eta n^{\frac{1}{4}}||x|-\sqrt{2n}|^{3/2}\right) ,&{}\quad \hbox {for}\quad \sqrt{2n}\le |x|\le (1+\tau )\sqrt{2n};\\ e^{-\xi x^2},&{}\quad \hbox {for}\quad (1+\tau )\sqrt{2n}\le |x|. \end{array}\right. \end{aligned}$$

(5)

A separate description of (4) is given in [1] due to Skovgaard. The unified and simplified form as (4) is stated in [7], in virtue of the relation [16], (1.1.52), (1.1.53)] of Hermite polynomials and Laguerre polynomials and a unified description [9], (2.2)] of Laguerre polynomials based on [8] and the table in [1, p. 699].

A direct consequence of (4) and (5) is

$$\begin{aligned} |{\mathcal {H}}_n(x)|\le An^{-\frac{1}{12}}, \end{aligned}$$

and ${\mathcal {H}}_n(x)$ attains this bound near the point $x=\sqrt{2n}$. But for most $x$ it has a much smaller bound as a multiple of $n^{-\frac{1}{4}}$. It is a very hard work to apply such a non-proportional property of the Hermite functions as Kanjin did in [5, 6], and certainly, it is also difficult to achieve the best result for related problems. However, if for $d\ge 2$, we denote by $\Phi _{\alpha }$, $\alpha \in {\mathbb {N}}^d$, the $d$-dimensional Hermite functions, namely,

$$\begin{aligned} \Phi _{\alpha }(x_1,\ldots ,x_d)={\mathcal {H}}_{\alpha _1}(x_1)\cdots {\mathcal {H}}_{\alpha _d}(x_d),\quad \alpha =(\alpha _1,\ldots ,\alpha _d), \end{aligned}$$

then there exists a constant $A>0$ independent of $n$ and $(x_1,\ldots ,x_d)$ such that (see [16, Lemma 3.2.2])

$$\begin{aligned} \sum _{|\alpha |=n}|\Phi _{\alpha }(x_1,\ldots ,x_d)|^2\le A(n+1)^{\frac{d}{2}-1}. \end{aligned}$$

(6)

Obviously this is not true for $d=1$. The bound in (6) has been used in research of various problems for $d\ge 2$, as in [16] for example; it is also the key in the proof of the inequalities of Hardy type in [10] for $d$-dimensional Hermite expansions for $d\ge 2$.

In order to prove the Hardy inequality (3), we shall follow a different approach, by evaluating the square integration of the Poisson integral associated to Hermite expansions of functions in $H^1({\mathbb {R}})$.

Indeed, we shall work with the generalized Hermite expansions of functions in $H^1({\mathbb {R}})$. If $\lambda >-1/2$, the generalized Hermite polynomials $H_n^{(\lambda )}(x)$ ($n\ge 0$) are defined by (see [3])

$$\begin{aligned} H_{2k}^{(\lambda )}(x)=&\left( \frac{k!}{\Gamma (k+\lambda +1/2)}\right) ^{1/2}L_k^{(\lambda -1/2)}(x^2),\end{aligned}$$

(7)

$$\begin{aligned} H_{2k-1}^{(\lambda )}(x)=&\left( \frac{(k-1)!}{\Gamma (k+\lambda +1/2)}\right) ^{1/2} xL_{k-1}^{(\lambda +1/2)}(x^2), \end{aligned}$$

(8)

where $L_n^{(\alpha )}(x)$ ($\alpha >-1$, $n\ge 0$) are the Laguerre polynomials determined by the orthogonal relation (see [15, 16])

$$\begin{aligned} \int _0^{\infty }e^{-x}x^{\alpha }L_n^{(\alpha )}(x)L_m^{(\alpha )}(x)dx=\frac{\Gamma (n+\alpha +1)}{n!}\delta _{mn}. \end{aligned}$$

The system $\left\{ H_n^{(\lambda )}(x)\right\}$ is orthonormal over $(-\infty ,+\infty )$ with respect to the weight $|x|^{2\lambda }e^{-x^2}$, and $H_n(x)=H_n^{(0)}(x)$ ($n\ge 0$) are the usual Hermite polynomials (up to constants).

The generalized Hermite functions ${\mathcal {H}}_n^{(\lambda )}(x)$ ($n\ge 0$) are given by

$$\begin{aligned} {\mathcal {H}}_n^{(\lambda )}(x)=e^{-\frac{x^2}{2}}|x|^{\lambda }H_n^{(\lambda )}(x), \end{aligned}$$

which are orthonormal over $(-\infty ,\infty )$ associated with Lebesgue measure. For a function $f\in L({\mathbb {R}})$, its generalized Hermite expansion is

$$\begin{aligned} f\sim \sum _{n=0}^{\infty }a_n^{(\lambda )}(f){\mathcal {H}}_n^{(\lambda )}(x),\qquad \qquad a_n^{(\lambda )}(f)=\int _{-\infty }^{\infty }f(t){\mathcal {H}}_n^{(\lambda )}(t)dt. \end{aligned}$$

(9)

In what follows we assume that $\lambda \ge 0$. Our main result is stated as follows.

Theorem 1.1

Let $\lambda \ge 0$. Then there exists a constant $A>0$ such that for all $f\in H^1({\mathbb {R}})$,

$$\begin{aligned} \sum _{n=1}^{\infty }n^{-\frac{3}{4}}|a_n^{(\lambda )}(f)|\le A\Vert f\Vert _{H^1({\mathbb {R}})}. \end{aligned}$$

(10)

The generalized Hermite polynomials $H_n^{(\lambda )}(x)$ ($n\ge 0$) were used in a Bose-like oscillator calculus in [11]; their further generalizations to the Dunkl setting in several variables can be found in [13] and those in the $A_{d-1}$ case are eigenfunctions of the Hamiltonian of the linear quantum Calogero-Moser-Sutherland model (with some modification) in ${\mathbb {R}}^d$ (see [13] and references therein). It would be interesting to extend various results on the usual Hermite expansions such as in [10] and others to orthogonal expansions associated to these generalizations. We remark that in [4], the Hermite functions were used in a description of Feichtinger’s space $S_0$.

Throughout the paper, $U\lesssim V$ means that $U\le cV$ for some positive constant $c$ independent of variables, functions, $n$, $k$, etc., but possibly dependent of the parameter $\lambda$.

2 Some Facts on the Poisson kernel

The Poisson kernel of the generalized Hermite polynomials $H_n^{(\lambda )}(x)$ ($n\ge 0$) is, for $0\le r<1$,

$$\begin{aligned} P^{(\lambda )}(r;x,y)=\sum _{n=0}^{\infty }r^nH_n^{(\lambda )}(x)H_n^{(\lambda )}(y). \end{aligned}$$

If we write $P^{(\lambda )}(r;x,y)$ into two parts, one being the summation for even $n$ and the other for odd $n$, then from (7), (8), and by [15] or [16], (1.1.47)], we have

$$\begin{aligned} P^{(\lambda )}(r;x,y)=\frac{(1-r^2)^{-\lambda -1/2}}{\Gamma (\lambda +1/2)} \exp \left( -\frac{r^2}{1-r^2}(x^2+y^2)\right) E_{\lambda }\left( \frac{2rxy}{1-r^2}\right) , \end{aligned}$$

where $E_{\lambda }(z)$ is the one-dimensional Dunkl kernel

$$\begin{aligned} E_\lambda (z)=j_{\lambda -1/2}(iz)+\frac{z}{2\lambda +1}j_{\lambda +1/2}(iz), \end{aligned}$$

and $j_{\alpha }(z)$ is the normalized Bessel function

$$\begin{aligned} j_{\alpha }(z)=2^{\alpha }\Gamma (\alpha +1)\frac{J_{\alpha }(z)}{z^{\alpha }} =\Gamma (\alpha +1)\sum _{n=0}^{\infty }\frac{(-1)^n(z/2)^{2n}}{n!\Gamma (n+\alpha +1)}. \end{aligned}$$

If $\lambda =0$, $E_0(z)=e^z$, and for $\lambda >0$, a Laplace-type representation of $E_\lambda (z)$ is (cf. [12, Lemma 2.1])

$$\begin{aligned} E_\lambda (z)=c_{\lambda }'\int _{-1}^1e^{zt}(1+t)(1-t^2)^{\lambda -1}dt, \quad c'_{\lambda }=\frac{\Gamma (\lambda +1/2)}{\Gamma (\lambda )\Gamma (1/2)}. \end{aligned}$$

(11)

By (), the Poisson kernel ${\mathcal {P}}^{(\lambda )}(r;x,y)=\sum _{n=0}^{\infty }r^n{\mathcal {H}}_n^{(\lambda )}(x){\mathcal {H}}_n^{(\lambda )}(y)$ of the generalized Hermite functions ${\mathcal {H}}_n^{(\lambda )}$ ($n\ge 0$) can be written as

$$\begin{aligned} {\mathcal {P}}^{(\lambda )}(r;x,y)=|xy|^{\lambda }\frac{(1-r^2)^{-\lambda -1/2}}{\Gamma (\lambda +1/2)} \exp \left( -\frac{1+r^2}{1-r^2}\cdot \frac{x^2+y^2}{2}\right) E_{\lambda }\left( \frac{2rxy}{1-r^2}\right) . \end{aligned}$$

The Poisson integral of the generalized Hermite expansion of a function $f\in L^1({\mathbb {R}})$ is defined by

$$\begin{aligned} f_{\lambda }(r;x)=\int _{-\infty }^{\infty }{\mathcal {P}}^{(\lambda )}(r;x,y)f(y)dy. \end{aligned}$$

(12)

For $\lambda =0$, ${\mathcal {P}}^{(0)}(r;x,y)$ is consistent with the Poisson kernel of the usual Hermite functions ${\mathcal {H}}_n$ ($n\ge 0$) (see [16])

$$\begin{aligned} {\mathcal {P}}(r;x,y)=\frac{(1-r^2)^{-\frac{1}{2}}}{\sqrt{\pi }} \exp \left( -\frac{2r(x-y)^2+(1-r)^2(x^2+y^2)}{2(1-r^2)}\right) . \end{aligned}$$

In order to give some necessary estimates of the Poisson kernel ${\mathcal {P}}^{(\lambda )}(r;x,y)$ we rewrite it into

$$\begin{aligned} {\mathcal {P}}^{(\lambda )}(r;x,y)=\frac{(1-r^2)^{-\lambda -1/2}}{\Gamma (\lambda +1/2)}|xy|^{\lambda }\psi _{r,x}(y)e^{-|\xi y|}E_{\lambda }(\xi y),\quad \xi =\frac{2rx}{1-r^2}, \end{aligned}$$

(13)

where

$$\begin{aligned} \psi _{r,x}(y)=\exp \left( -\frac{1+r^2}{1-r^2}\frac{x^2+y^2}{2}+\frac{2r|xy|}{1-r^2}\right) . \end{aligned}$$

We shall need several preliminary lemmas.

Lemma 2.1

For $y\ne 0$,

• (i) $\left| e^{-|\xi y|}E_{\lambda }(\xi y)\right| \lesssim (1+|\xi y|)^{-\lambda }$;

• (ii) $\left| \frac{\partial }{\partial y}\left( e^{-|\xi y|}E_{\lambda }(\xi y)\right) \right| \lesssim |\xi |(1+|\xi y|)^{-\lambda -1}$.

Proof

If $\lambda =0$, both (i) and (ii) are trivial. In what follows we assume that $\lambda >0$. From (11) we have

$$\begin{aligned} \left| e^{-|\xi y|}E_{\lambda }(\xi y)\right|&\le 2c_{\lambda }'\int _{-1}^1e^{-|\xi y|(1-|t|)}(1-t^2)^{\lambda -1}dt \lesssim \int _{0}^1e^{-|\xi y|(1-t)}(1-t)^{\lambda -1}dt\\&=|\xi y|^{-\lambda }\int _0^{|\xi y|}e^{-s}s^{\lambda -1}ds, \end{aligned}$$

which is bounded by a multiple of $(1+|\xi y|)^{-\lambda }$, since $\int _0^{A}e^{-s}s^{\lambda -1}ds\asymp \left( A/(A+1)\right) ^{\lambda }$.

Furthermore, from (11) we have

$$\begin{aligned} \frac{\partial }{\partial y}\left( e^{-|\xi y|}E_{\lambda }(\xi y)\right)&=-c_{\lambda }'\xi \int _{-1}^1e^{-\xi y(1-t)}(1+t)^{\lambda }(1-t)^{\lambda }dt\quad \hbox {for}\quad \xi y>0;\\&=c_{\lambda }'\xi \int _{-1}^1e^{\xi y(1-t)}(1+t)^{\lambda -1}(1-t)^{\lambda +1}dt\quad \,\, \hbox {for}\quad \xi y<0. \end{aligned}$$

In the both cases we split the integral into two parts as $\int _{-1}^1=\int _{-1}^0+\int _0^1$, to get

$$\begin{aligned} \left| \frac{\partial }{\partial y}\left( e^{-|\xi y|}E_{\lambda }(\xi y)\right) \right|&\lesssim |\xi |\left[ e^{-|\xi y|}+\int _{0}^1e^{-|\xi y|(1-t)}(1-t)^{\lambda }dt\right] \\&=|\xi |\left[ e^{-|\xi y|}+|\xi y|^{-\lambda -1}\int _0^{|\xi y|}e^{-s}s^{\lambda }ds\right] \end{aligned}$$

which implies the desired estimate in part (ii), since $\int _0^{A}e^{-s}s^{\lambda }ds\asymp \left( A/(A+1)\right) ^{\lambda +1}$. $\square$

Lemma 2.2

• (i) For $0\le r<1$,

  $$\begin{aligned} \psi _{r,x}(y)=\exp \left( -\frac{r(|x|-|y|)^2}{1-r^2}\right) \exp \left( -\frac{1-r}{1+r}\frac{x^2+y^2}{2}\right) ; \end{aligned}$$

  (14)

• (ii) for $0\le r\le \frac{1}{2}$, $\psi _{r,x}(y)\lesssim \exp \left( -c_1(x^2+y^2)\right)$, and for $\frac{1}{2}\le r<1$, $\psi _{r,x}(y)\lesssim \exp \left( -c_2\frac{(|x|-|y|)^2}{1-r}\right)$, where $c_1,c_2>0$ are independent of $x,y$ and $r$.

Part (i) is obvious and implies part (ii) immediately.

Lemma 2.3

For $x,y\in (-\infty ,\infty )$ and $r\in [0,1)$,

$$\begin{aligned} \frac{(1-r)^{1/2}|x|}{1-r+|xy|}\lesssim 1+\frac{\left| |x|-|y|\right| }{(1-r)^{1/2}}. \end{aligned}$$

Indeed, for $|y|\le |x|/2$ the left hand side is bounded by $|x|/(1-r)^{1/2}\le 2\left| |x|-|y|\right| /(1-r)^{1/2}$, and for $|y|\ge |x|/2$, by $|x|/\left( 2|xy|^{1/2}\right) \le 1$.

Proposition 2.4

• (i) If $0\le r\le \frac{1}{2}$, then for $y\ne 0$,

  $$\begin{aligned} \left| \frac{\partial }{\partial y}{\mathcal {P}}^{(\lambda )}(r;x,y)\right| \lesssim |x|^{\lambda }|y|^{\lambda -1}\left( \lambda +|xy|+|y|^2\right) \exp \left( -c_1(x^2+y^2)\right) ;\nonumber \\ \end{aligned}$$

  (15)

• (ii) if $\frac{1}{2}\le r<1$, then for $y\ne 0$,

  $$\begin{aligned} \left| \frac{\partial }{\partial y}{\mathcal {P}}^{(\lambda )}(r;x,y)\right| \lesssim \lambda \frac{(1-r)^{-1/2}|x|^{\lambda }|y|^{\lambda -1}}{(1-r+|xy|)^{\lambda }}\psi _{r,x}(y) +(1-r)^{-1}\psi _{r,x}(y)^{\frac{1}{2}}.\nonumber \\ \end{aligned}$$

  (16)

Proof

For $y\ne 0$, from (13) we have

$$\begin{aligned} \frac{\partial }{\partial y}{\mathcal {P}}^{(\lambda )}(r;x,y)=&\frac{(1-r^2)^{-\lambda -1/2}}{\Gamma (\lambda +1/2)}\left( U_1(x,y)+U_2(x,y)\right) , \end{aligned}$$

where

$$\begin{aligned}&U_1(x,y)=\lambda |x|^{\lambda }|y|^{\lambda -2}y\left( \psi _{r,x}(y)e^{-|\xi y|}E_{\lambda }(\xi y)\right) ,\\ {}&U_2(x,y)=|xy|^{\lambda }\left( \psi _{r,x}(y)e^{-|\xi y|}E_{\lambda }(\xi y)\right) '_y. \end{aligned}$$

For $0\le r\le \frac{1}{2}$, it follows from Lemmas 2.1(i) and 2.2(ii) that $(1-r^2)^{-\lambda -1/2}U_1(x,y)$ is bounded by a multiple of $\lambda |x|^{\lambda }|y|^{\lambda -1}\exp \left( -c_1(x^2+y^2)\right)$, and for $\frac{1}{2}\le r<1$, by a multiple of

$$\begin{aligned} \lambda \frac{(1-r)^{-\lambda -1/2}|x|^{\lambda }|y|^{\lambda -1}}{(1+|\xi y|)^{\lambda }}\psi _{r,x}(y) \lesssim \lambda \frac{(1-r)^{-1/2}|x|^{\lambda }|y|^{\lambda -1}}{(1-r+|xy|)^{\lambda }}\psi _{r,x}(y). \end{aligned}$$

In order to verify that $(1-r^2)^{-\lambda -1/2}U_2(x,y)$ has the desired estimates as in (15) and (16), we shall show that for $0\le r\le \frac{1}{2}$,

$$\begin{aligned} \left| \left( \psi _{r,x}(y)e^{-|\xi y|}E_{\lambda }(\xi y)\right) '_y\right| \lesssim (|x|+|y|)\exp \left( -c_1(x^2+y^2)\right) ; \end{aligned}$$

(17)

and for $\frac{1}{2}\le r<1$,

$$\begin{aligned}&\left| \left( \psi _{r,x}(y)e^{-|\xi y|}E_{\lambda }(\xi y)\right) '_y\right| \nonumber \\ {}&\quad \lesssim \frac{\psi _{r,x}(y)}{(1-r)^{1-\lambda }} \left[ \frac{\left| |x|-|y|\right| +(1-r)^2|y|}{(1-r+|xy|)^{\lambda }}+\frac{(1-r)|x|}{(1-r+|xy|)^{\lambda +1}}\right] . \end{aligned}$$

(18)

Indeed, since from (14),

$$\begin{aligned}&\left( \psi _{r,x}(y)e^{-|\xi y|}E_{\lambda }(\xi y)\right) '_y\\&\quad =\left[ \frac{2r(|x|-|y|)(\hbox {sgn}\,y){-}(1{-}r)^2y}{1{-}r^2}\left( e^{-|\xi y|}E_{\lambda }(\xi y)\right) {+}\left( e^{-|\xi y|}E_{\lambda }(\xi y)\right) '_y\right] \psi _{r,x}(y), \end{aligned}$$

by Lemma 2.1 we have

$$\begin{aligned} \left| \left( \psi _{r,x}(y)e^{-|\xi y|}E_{\lambda }(\xi y)\right) '_y\right| \lesssim \left[ \frac{\left| |x|-|y|\right| +(1-r)^2|y|}{(1-r)(1+|\xi y|)^{\lambda }} +\frac{|\xi |}{(1+|\xi y|)^{\lambda +1}}\right] \psi _{r,x}(y). \end{aligned}$$

Thus (17) and (18) follow immediately.

For $0\le r\le \frac{1}{2}$, from (17) it is obvious that

$$\begin{aligned} \left| (1-r^2)^{-\lambda -1/2}U_2(x,y)\right| \lesssim |xy|^{\lambda }(|x|+|y|)\exp \left( -c_1(x^2+y^2)\right) ; \end{aligned}$$

and for $\frac{1}{2}\le r<1$, from (18) we have

$$\begin{aligned} \left| (1-r^2)^{-\lambda -1/2}U_2(x,y)\right| \lesssim \frac{\psi _{r,x}(y)}{1-r} \left[ \frac{\left| |x|-|y|\right| }{(1{-}r)^{1/2}}{+}(1-r)^{3/2}|y|+\frac{(1-r)^{1/2}|x|}{1-r+|xy|}\right] . \end{aligned}$$

By Lemma 2.3, it is easy to see that the expression in the brackets above is bounded by a multiple of $\psi _{r,x}(y)^{-1/2}$, and so $\left| (1-r^2)^{-\lambda -1/2}U_2(x,y)\right| \lesssim \psi _{r,x}(y)^{1/2}/(1-r)$. The proof of the proposition is finished. $\square$

3 Proof of the Main Result

The following theorem is crucial in the proof of Theorem 1.1.

Theorem 3.1

Assume that $\lambda \ge 0$. There is a positive constant $A$ such that for all $f\in H^1({\mathbb {R}})$,

$$\begin{aligned} \int _0^1(1-r)^{-\frac{3}{4}}\left( \int _{-\infty }^{\infty }|f_{\lambda }(r;y)|^2dy\right) ^{\frac{1}{2}}dr\le A\Vert f\Vert _{H^1({\mathbb {R}})}, \end{aligned}$$

(19)

where $f_{\lambda }(r;y)$ is the Poisson integral (12) associated to the generalized Hermite expansion of $f$.

The proof of Theorem 3.1 is based upon some norm estimates stated below.

Lemma 3.2

• (i) If $\lambda =0$ or $\lambda >1$, then for $y,\bar{y}\in (-\infty ,\infty )$,

  $$\begin{aligned} \left( \int _{-\infty }^{\infty } \left| {\mathcal {P}}^{(\lambda )}(r;x,y)-{\mathcal {P}}^{(\lambda )}(r;x,\bar{y})\right| ^2dx\right) ^{\frac{1}{2}} \lesssim \frac{|y-\bar{y}|}{(1-r)^{3/4}}; \end{aligned}$$

  (20)

• (ii) If $0<\lambda \le 1$, then for $y,\bar{y}\in (-\infty ,\infty )$,

  $$\begin{aligned} \left( \int _{-\infty }^{\infty } \left| {\mathcal {P}}^{(\lambda )}(r;x,y)-{\mathcal {P}}^{(\lambda )}(r;x,\bar{y})\right| ^2dx\right) ^{\frac{1}{2}} \lesssim \frac{|y{-}\bar{y}|^{\lambda }}{(1{-}r)^{(2\lambda {+}1)/4}} \left( 1{+}\frac{|y{-}\bar{y}|}{(1{-}r)^{1/2}}\right) . \end{aligned}$$

Proof

For $\lambda >1$ and $y>\bar{y}$ we have

$$\begin{aligned} \left\| {\mathcal {P}}^{(\lambda )}(r;\cdot ,y)-{\mathcal {P}}^{(\lambda )}(r;\cdot ,\bar{y})\right\| _{L^2({\mathbb {R}})}&=\left\| \int _{\bar{y}}^y\frac{\partial }{\partial z}{\mathcal {P}}^{(\lambda )}(r;\cdot ,z)dz\right\| _{L^2({\mathbb {R}})}\\&\le \int _{\bar{y}}^y\left\| \frac{\partial }{\partial z}{\mathcal {P}}^{(\lambda )}(r;\cdot ,z)\right\| _{L^2({\mathbb {R}})}dz. \end{aligned}$$

If $0\le r\le \frac{1}{2}$, then by Proposition 2.4(i), $\left\| \frac{\partial }{\partial z}{\mathcal {P}}^{(\lambda )}(r;\cdot ,z)\right\| _{L^2({\mathbb {R}})}\lesssim 1$, so that

$$\begin{aligned} \left\| {\mathcal {P}}^{(\lambda )}(r;\cdot ,y)-{\mathcal {P}}^{(\lambda )}(r;\cdot ,\bar{y})\right\| _{L^2({\mathbb {R}})} \lesssim |y-\bar{y}|\lesssim \frac{|y-\bar{y}|}{(1-r)^{3/4}}. \end{aligned}$$

Since for $\lambda >1$ and $\frac{1}{2}\le r<1$, by Lemma 2.3 we have

$$\begin{aligned} \frac{(1-r)^{-1/2}|x|^{\lambda }|y|^{\lambda -1}}{(1-r+|xy|)^{\lambda }} \le \frac{(1-r)^{-1/2}|x|}{1-r+|xy|}\lesssim \frac{1}{1-r}\left( 1+\frac{\left| |x|-|y|\right| }{(1-r)^{1/2}}\right) \lesssim \frac{\psi _{r,x}(y)^{-\frac{1}{2}}}{1-r}, \end{aligned}$$

inserting this into (16) yields that $\left| \frac{\partial }{\partial y}{\mathcal {P}}^{(\lambda )}(r;x,y)\right| \lesssim (1-r)^{-1}\psi _{r,x}(y)^{\frac{1}{2}}$ and hence

$$\begin{aligned} \left\| {\mathcal {P}}^{(\lambda )}(r;\cdot ,y)-{\mathcal {P}}^{(\lambda )}(r;\cdot ,\bar{y})\right\| _{L^2({\mathbb {R}})} \le \frac{1}{1-r}\int _{\bar{y}}^y\left\| \psi _{r,x}(z)^{\frac{1}{2}}\right\| _{L^2({\mathbb {R}})}dz \lesssim \frac{|y-\bar{y}|}{(1-r)^{3/4}}. \end{aligned}$$

This proves part (i) for $\lambda >1$.

If $\lambda =0$, by Proposition 2.4 $\left| \frac{\partial }{\partial y}{\mathcal {P}}^{(0)}(r;x,y)\right|$ is bounded by a multiple of $(|x|+|y|)e^{-c_1(x^2+y^2)}$ for $0\le r\le \frac{1}{2}$, and $(1-r)^{-1}\psi _{r,x}(y)^{\frac{1}{2}}$ for $\frac{1}{2}\le r<1$. Hence the above process still works for $\lambda =0$. This completes the proof of part (i).

Now we turn to the proof of part (ii). For $0<\lambda \le 1$ and $y>\bar{y}$, we write

$$\begin{aligned}&{\mathcal {P}}^{(\lambda )}(r;x,y)-{\mathcal {P}}^{(\lambda )}(r;x,\bar{y})\\&\quad = \frac{(1-r^2)^{-\lambda -1/2}}{\Gamma (\lambda +1/2)}\left[ \int _{\bar{y}}^y U_2(x,z)dz +|x|^{\lambda }(|y|^{\lambda }-|\bar{y}|^{\lambda })\psi _{r,x}(y)e^{-|\xi y|}E_{\lambda }(\xi y)\right. \\&\quad \quad +\left. \int _{\bar{y}}^y|x|^{\lambda }(|\bar{y}|^{\lambda }-|z|^{\lambda }) \left( \psi _{r,x}(z)e^{-|\xi z|}E_{\lambda }(\xi z)\right) '_zdz\right] , \end{aligned}$$

where $U_2(x,z)$ is defined as the same as in the proof of Proposition 2.4. There we have shown that (for all $\lambda \ge 0$) $(1-r^2)^{-\lambda -1/2}U_2(x,z)$ is bounded by a multiple of $|xz|^{\lambda }(|x|+|z|)\exp \left( -c_1(x^2+z^2)\right)$ for $0\le r\le 1/2$, and of $\psi _{r,x}(z)^{1/2}/(1-r)$ for $1/2\le r<1$. From the proof of part (i) above it follows that

$$\begin{aligned} \left\| {\mathcal {P}}^{(\lambda )}(r;\cdot ,y)-{\mathcal {P}}^{(\lambda )}(r;\cdot ,\bar{y})\right\| _{L^2({\mathbb {R}})} \lesssim&\frac{|y-\bar{y}|}{(1-r)^{3/4}}+V_1+V_2, \end{aligned}$$

(21)

where

$$\begin{aligned} V_1&=\frac{|y-\bar{y}|^{\lambda }}{(1-r)^{\lambda +1/2}} \left( \int _{-\infty }^{\infty }\left| |x|^{\lambda }\psi _{r,x}(y)e^{-|\xi y|}E_{\lambda }(\xi y)\right| ^2dx\right) ^{1/2},\\ V_2&=\frac{|y-\bar{y}|^{\lambda }}{(1-r)^{\lambda +1/2}}\int _{\bar{y}}^y\left( \int _{-\infty }^{\infty }\left| |x|^{\lambda } \left( \psi _{r,x}(z)e^{-|\xi z|}E_{\lambda }(\xi z)\right) '_z\right| ^2dx\right) ^{1/2}dz. \end{aligned}$$

If $0\le r\le \frac{1}{2}$, then by Lemmas 2.1(i) and 2.2(ii),

$$\begin{aligned} V_1&\lesssim |y-\bar{y}|^{\lambda } \left( \int _{-\infty }^{\infty }|x|^{2\lambda }\exp \left( -2c_1(x^2+y^2)\right) dx\right) ^{1/2}\nonumber \\&\lesssim |y-\bar{y}|^{\lambda }\lesssim \frac{|y-\bar{y}|^{\lambda }}{(1-r)^{(2\lambda +1)/4}}; \end{aligned}$$

(22)

and from (17),

$$\begin{aligned} V_2&\lesssim |y-\bar{y}|^{\lambda } \int _{\bar{y}}^y\left( \int _{-\infty }^{\infty }|x|^{2\lambda }(|x|+|z|)^2\exp \left( -2c_1(x^2+z^2)\right) dx\right) ^{1/2}dz\nonumber \\&\lesssim |y-\bar{y}|^{\lambda +1}\lesssim \frac{|y-\bar{y}|^{\lambda +1}}{(1-r)^{(2\lambda +3)/4}}. \end{aligned}$$

(23)

If $\frac{1}{2}\le r<1$, then by Lemmas 2.1(i), 2.2(ii) and 2.3,

$$\begin{aligned} V_1&\lesssim \frac{|y-\bar{y}|^{\lambda }}{(1-r)^{1/2}} \left( \int _{-\infty }^{\infty }\left| \psi _{r,x}(y)\frac{|x|^{\lambda }}{(1-r+|xy|)^{\lambda }}\right| ^2dx\right) ^{1/2} \nonumber \\&\lesssim \frac{|y-\bar{y}|^{\lambda }}{(1-r)^{(\lambda +1)/2}} \left( \int _{-\infty }^{\infty }\left( 1+\frac{\left| |x|-|y|\right| }{(1-r)^{1/2}}\right) ^{2\lambda } \exp \left( -2c_2\frac{(|x|-|y|)^2}{1-r}\right) dx\right) ^{1/2}\nonumber \\&\lesssim \frac{|y-\bar{y}|^{\lambda }}{(1-r)^{(2\lambda +1)/4}}; \end{aligned}$$

(24)

and by (18), Lemmas 2.2(ii) and 2.3,

$$\begin{aligned} V_2&\lesssim \frac{|y-\bar{y}|^{\lambda }}{(1-r)^{(\lambda +2)/2}} \int _{\bar{y}}^y\left( \int _{-\infty }^{\infty }\left[ \left( \frac{(1-r)^{1/2}|x|}{1-r+|xz|}\right) ^{\lambda } \left( \frac{\left| |x|-|z|\right| }{(1-r)^{1/2}}+(1-r)^{\frac{3}{2}}|z|\right) \right. \right. \nonumber \\&\left. \left. \qquad +\left( \frac{(1-r)^{1/2}|x|}{1-r+|xz|}\right) ^{\lambda +1}\right] ^2\psi _{r,x}(z)^2dx\right) ^{1/2}dz\nonumber \\&\lesssim \frac{|y-\bar{y}|^{\lambda }}{(1-r)^{(\lambda +2)/2}} \int _{\bar{y}}^y\left( \int _{-\infty }^{\infty }\left[ \left( 1+\frac{\left| |x|-|z|\right| }{(1-r)^{1/2}}\right) ^{\lambda } \left( \frac{\left| |x|-|z|\right| }{(1-r)^{1/2}}+(1-r)^{\frac{3}{2}}|z|\right) \right. \right. \nonumber \\&\left. \left. \qquad +\left( 1+\frac{\left| |x|-|z|\right| }{(1-r)^{1/2}}\right) ^{\lambda +1}\right] ^2\psi _{r,x}(z)^2dx\right) ^{1/2}dz\nonumber \\&\lesssim \frac{|y-\bar{y}|^{\lambda }}{(1-r)^{(\lambda +2)/2}} \int _{\bar{y}}^y\left( \int _{-\infty }^{\infty }\psi _{r,x}(z)dx\right) ^{1/2}dz\nonumber \\&\lesssim \frac{|y-\bar{y}|^{\lambda +1}}{(1-r)^{(2\lambda +3)/4}}. \end{aligned}$$

(25)

Collecting the estimates of (22), (23), (24) and (25) into (21), we finish the proof of part (ii) of the lemma. $\square$

Now we come to the proof of Theorem 3.1.

Let us first recall the atom characterization of the Hardy space $H^1({\mathbb {R}})$ (cf. [14, 18]). An $H^1$ atom is a measurable function $a$ on ${\mathbb {R}}$ satisfying

• (i) $\hbox {supp}\, a\subseteq I$ for some interval $I\subset {\mathbb {R}}$;

• (ii) $\Vert a\Vert _{L^{\infty }}\le |I|^{-1}$, $|I|$ denoting the length of $I$;

• (iii) $\int _{-\infty }^{\infty }a(x)dx=0$.

For a function $f\in H^1({\mathbb {R}})$, there are a sequence $\{a_k\}$ of $H^1$ atoms and a sequence $\{\lambda _k\}$ of complex numbers satisfying $\sum |\lambda _k|<\infty$, such that $f=\sum _{k=1}^{\infty }\lambda _k a_k$ and

$$\begin{aligned} A_1\Vert f\Vert _{H^1({\mathbb {R}})}\le \sum |\lambda _k|\le A_2\Vert f\Vert _{H^1({\mathbb {R}})}, \end{aligned}$$

where $A_1,A_2$ are positive constants independent of $f$.

In order to prove Theorem 3.1, it suffices to verify the inequality (19) for each $H^1$ atom $f=a$, namely, there exists a absolute constant $A>0$, such that

$$\begin{aligned} \int _0^1(1-r)^{-\frac{3}{4}}\Vert a_{\lambda }(r;\cdot )\Vert _{L^2({\mathbb {R}})}dr\le A \end{aligned}$$

(26)

holds for all $H^1$ atoms $a$ satisfying (i), (ii) and (iii) above, where $a_{\lambda }(r;y)$ is the Poisson integral (12) of $a$ associated to its generalized Hermite expansion.

By Parseval’s inequality, from (i) and (ii) we have

$$\begin{aligned} \Vert a_{\lambda }(r;\cdot )\Vert _{L^2({\mathbb {R}})}\le \Vert a\Vert _{L^2({\mathbb {R}})}\le |I|^{-\frac{1}{2}}. \end{aligned}$$

(27)

(27) is useful for atoms with larger supports, but we also need a more accurate estimate for atoms with smaller supports.

For an atom $a$ satisfying (i), (ii) and (iii), let $\bar{y}$ be the left endpoint of the interval $I$. Applying the cancelation property (iii) and Minkowski’s inequality, we have

$$\begin{aligned} \Vert a_{\lambda }(r;\cdot )\Vert _{L^2({\mathbb {R}})}&=\left( \int _{-\infty }^{\infty } \left[ \int _I\left( {\mathcal {P}}^{(\lambda )}(r;x,y)-{\mathcal {P}}^{(\lambda )}(r;x,\bar{y})\right) a(y)dy\right] ^{2}dx \right) ^{\frac{1}{2}}\nonumber \\&\le \int _I|a(y)|\left( \int _{-\infty }^{\infty } \left| {\mathcal {P}}^{(\lambda )}(r;x,y)-{\mathcal {P}}^{(\lambda )}(r;x,\bar{y})\right| ^2dx\right) ^{1/2}dy. \end{aligned}$$

By means of Lemma 3.2 we obtain

$$\begin{aligned} \Vert a_{\lambda }(r;\cdot )\Vert _{L^2({\mathbb {R}})}\lesssim \frac{|I|}{(1-r)^{3/4}} +\frac{|I|^{\lambda }}{(1-r)^{(2\lambda +1)/4}}+\frac{|I|^{\lambda +1}}{(1-r)^{(2\lambda +3)/4}}. \end{aligned}$$

(28)

If $\lambda =0$ or $\lambda >1$, the first term on the right hand side appears only.

If $|I|\ge 1$, from (27) it is trivial that

$$\begin{aligned} \int _0^1(1-r)^{-\frac{3}{4}}\Vert a_{\lambda }(r;\cdot )\Vert _{L^2({\mathbb {R}})}dr\le 4|I|^{-\frac{1}{2}}\le 4; \end{aligned}$$

and if $|I|<1$, we split the integral into two parts, over $[1-|I|^2,1)$ and $[0,1-|I|^2]$ respectively, and apply (27) and (28) to obtain

$$\begin{aligned}&\int _0^1(1-r)^{-\frac{3}{4}}\Vert a_{\lambda }(r;\cdot )\Vert _{L^2({\mathbb {R}})}dr\lesssim \int _{1-|I|^2}^1|I|^{-1/2}(1-r)^{-3/4}dr\\&\quad +\int _0^{1-|I|^2}\left( \frac{|I|}{(1-r)^{3/2}} +\frac{|I|^{\lambda }}{(1-r)^{(\lambda +2)/2}}+\frac{|I|^{\lambda +1}}{(1-r)^{(\lambda +3)/2}}\right) dr. \end{aligned}$$

It is easy to see that the values of the two integrals above are independent of $|I|<1$. This proves (26), and hence Theorem 3.1.

Proof of Theorem 1.1

By Hölder’s inequality one has

$$\begin{aligned} \sum _{n=0}^{\infty }r^{2n}|a_n^{(\lambda )}(f)|&\le \left( \sum _{n=0}^{\infty }r^{2n}\right) ^{1/2}\left( \sum _{n=0}^{\infty }r^{2n}|a_n^{(\lambda )}(f)|^{2}\right) ^{1/2}\\&\le (1-r)^{-1/2}\Vert f_{\lambda }(r;\cdot )\Vert _{L^2}. \end{aligned}$$

Multiplying $(1-r)^{-1/4}$ on both sides and taking integration over $[0,1)$ respectively, then by Theorem 3.1 we get

$$\begin{aligned} \sum _{n=0}^{\infty }B(2n+1,3/4)|a_n^{(\lambda )}(f)|\le A\Vert f\Vert _{H^1}, \end{aligned}$$

where $B(\alpha ,\beta )$ denotes the beta function. Finally, the Hardy inequality (10) follows from the fact that $B(2n+1,3/4)\sim (n+1)^{-3/4}$. The proof of Theorem 1.1 is completed.$\square$