A Hardy Inequality for Ultraspherical Expansions with an Application to the Sphere

Abstract

We prove a Hardy inequality for ultraspherical expansions by using a proper ground state representation. From this result we deduce some uncertainty principles for this kind of expansions. Our result also implies a Hardy inequality on spheres with a potential having a double singularity.

1 Introduction and Main Result

For $d\ge 3$, the classical Hardy inequality states that

$$\begin{aligned} \frac{(d-2)^2}{4}\int _{\mathbb {R}^d}\frac{u^2(x)}{|x|^2}\, dx\le \int _{\mathbb {R}^d}|\nabla u(x)|^2\, dx. \end{aligned}$$

(1)

Due to its applicability, there is an extensive literature about the topic (see the references in [16]) covering many extensions of this estimate in several and different directions. We are interested in one involving the fractional powers of the Laplacian. We can rewrite (1) as

$$\begin{aligned} \frac{(d-2)^2}{4}\int _{\mathbb {R}^d}\frac{u^2(x)}{|x|^2}\, dx\le \int _{\mathbb {R}^d} u(x) (-\Delta u(x)) \, dx \end{aligned}$$

and, taking the fractional Laplacian $(-\Delta )^\sigma$ defined by $\widehat{(-\Delta )^\sigma u}=|\cdot |^{2\sigma } \widehat{u}$, a natural extension is the inequality

$$\begin{aligned} C_{\sigma ,d}\int _{\mathbb {R}^d}\frac{u^2(x)}{|x|^{2\sigma }}\, dx\le \int _{\mathbb {R}^d} u(x) (-\Delta )^{\sigma } u(x) \, dx, \end{aligned}$$

(2)

for which the sharp constant $C_{\sigma ,d}$ is well known (see [3, 20]).

From (2), we deduce the positivity (in a distributional sense) of the operator

$$\begin{aligned} (-\Delta )^\sigma -\frac{C_{\sigma ,d}}{|\cdot |^{2\sigma }}. \end{aligned}$$

Our target is to provide a Hardy inequality like (2) related to ultraspherical expansions and apply it to prove the positivity of certain operator on the sphere with a potential having singularities in both poles of the sphere.

Let $C_n^\lambda (x)$ be the ultraspherical polynomial of degree n and order $\lambda >-1/2$. We consider $c_n^\lambda (x)=d_n^{-1}C_n^\lambda (x)$ with

$$\begin{aligned} d_n^2= \int _{-1}^{1}\left( C_n^\lambda (x)\right) ^2\,d\mu _\lambda (x),\qquad d\mu _\lambda (x)=(1-x^2)^{\lambda -1/2}\, dx. \end{aligned}$$

The sequence of polynomials $\{c_n^\lambda \}_{n\ge 0}$ forms an orthonormal basis of the space $L_\lambda ^2:=L^2((-1,1),d\mu _\lambda )$. For each $c_n^\lambda$, it holds that $\mathcal {L}_\lambda c_n^\lambda =-(n+\lambda )^2 c_n^\lambda$, where

$$\begin{aligned} \mathcal {L}_\lambda =(1-x^2)\frac{d^2}{dx^2}-(2\lambda +1)x\frac{d}{dx}-\lambda ^2. \end{aligned}$$

The ultraspherical expansion of each appropriate function f defined in $(-1,1)$ is given by

$$\begin{aligned} f\longmapsto \sum _{n=0}^\infty a_n^\lambda (f)c_n^\lambda , \end{aligned}$$

where $a_n^\lambda (f)$ is the n-th Fourier coefficient of f respect to $\{c_n^\lambda \}_{n\ge 0}$, i.e.,

$$\begin{aligned} a_n^\lambda (f)=\int _{-1}^{1}f(y)c_n^\lambda (y)\,d\mu _\lambda (y). \end{aligned}$$

The fractional powers of the operator $\mathcal {L}_\lambda$ are defined by

$$\begin{aligned} (-\mathcal {L}_\lambda )^{\sigma /2}f=\sum _{n=0}^\infty (n+\lambda )^{\sigma } a_n^\lambda (f)c_n^\lambda , \qquad \sigma >0. \end{aligned}$$

This operator should be the natural candidate to prove a Hardy type inequality for the ultraspherical expansion but, however, it is not the most appropriate in this setting. We have to consider another one with an analogous behaviour to $(-\mathcal {L}_\lambda )^{\sigma /2}$, in order to deduce some results on the sphere. For each $\sigma >0$ we define (spectrally) the operator

$$\begin{aligned} A_\sigma ^{\lambda } =\frac{\Gamma (\sqrt{-\mathcal {L}_\lambda }+\frac{1+\sigma }{2})}{\Gamma (\sqrt{-\mathcal {L}_\lambda }+\frac{1-\sigma }{2})}. \end{aligned}$$

Then for f defined on the interval $(-1,1)$

$$\begin{aligned} A_\sigma ^{\lambda } f(x)=\sum _{n=0}^{\infty } \frac{\Gamma (n+\lambda +\frac{1+\sigma }{2})}{\Gamma (n+\lambda +\frac{1-\sigma }{2})}a_n^\lambda (f)c_n^\lambda (x). \end{aligned}$$

Note that

$$\begin{aligned} \frac{\Gamma (n+\lambda +\frac{1+\sigma }{2})}{\Gamma (n+\lambda +\frac{1-\sigma }{2})}\simeq (n+\lambda )^{\sigma }, \end{aligned}$$

(3)

then the behaviour of $(-\mathcal {L}_\lambda )^{\sigma /2}$ and $A_\sigma ^\lambda$ is similar. The natural Sobolev space to analyse Hardy type inequalities is

$$\begin{aligned} H^{\sigma }_\lambda =\Big \{f\in L_\lambda ^2: \Vert f\Vert _{H^{\sigma }_\lambda }:=\Big (\sum _{n=0}^\infty (n+\lambda )^{\sigma }(a_n^\lambda (f))^2\Big )^{1/2}<\infty \Big \}. \end{aligned}$$

We have to note that $H^\sigma _\lambda$ is equivalent to the space $\mathcal {L}_{\lambda ,\sigma }^2$ introduced in [5].

With the previous notation our Hardy inequality for ultraspherical expansions is given in the following result.

Theorem 1

Let $\lambda >0$ and $0<\sigma <1$. Then for $u\in H_\lambda ^\sigma$

$$\begin{aligned} Q_{\sigma ,\lambda }\int _{-1}^{1}\frac{u^2(x)}{(1-x^2)^{\sigma /2}}\,d\mu _\lambda (x)\le \int _{-1}^{1}u(x)A_\sigma ^{\lambda } u(x)\,d\mu _\lambda (x), \end{aligned}$$

(4)

where

$$\begin{aligned} Q_{\sigma ,\lambda }=2^\sigma \frac{\Gamma (\frac{\lambda }{2}+\frac{1+\sigma }{4})^2}{\Gamma (\frac{\lambda }{2}+\frac{1-\sigma }{4})^2}. \end{aligned}$$

(5)

Inequality (4) can be rewritten in terms of the Fourier coefficients

$$\begin{aligned} Q_{\sigma ,\lambda }\int _{-1}^{1}\frac{u^2(x)}{(1-x^2)^{\sigma /2}}\, d\mu _\lambda (x)\le \sum _{n=0}^\infty \frac{\Gamma (n+\lambda +\frac{1+\sigma }{2})}{\Gamma (n+\lambda +\frac{1-\sigma }{2})}(a_n^\lambda (u))^2, \end{aligned}$$

(6)

which is a kind of Pitt inequality for the ultraspherical expansions (for other Pitt inequalities see [4, 11]). Note that for the right hand side of (4) we have, by (3),

$$\begin{aligned} \int _{-1}^{1}u(x)A_\sigma ^{\lambda } u(x)\,d\mu _\lambda (x)=\sum _{n=0}^\infty \frac{\Gamma (n+\lambda +\frac{1+\sigma }{2})}{\Gamma (n+\lambda +\frac{1-\sigma }{2})}(a_n^\lambda (u))^2 \simeq \Vert u\Vert _{H^{\sigma }_\lambda }^2, \end{aligned}$$

so the space $H_\lambda ^\sigma$ is the adequated one.

The proof of Theorem 1 will be a consequence of a proper ground state representation in our setting, analogous to the given one in the Euclidean case in [9]. Following the ideas in that paper, we can see that the constant $Q_{\sigma ,\lambda }$ is sharp but not achieved. Similar ideas have been recently exploited in [7, 16].

From (4), by using Cauchy–Schwarz inequality, we can obtain a Heisenberg type uncertainty principle as it was done for the sublaplacian of the Heisenberg group in [10], and for the fractional powers of the same sublaplacian in [16].

Corollary 2

Let $\lambda >0$ and $0<\sigma <1$. Then for $u\in H_\lambda ^\sigma$

$$\begin{aligned} Q_{\sigma ,\lambda }\left( \int _{-1}^{1} u^2(x)\,d\mu _\lambda (x)\right) ^{2}\le & {} \int _{-1}^{1}u^2(x)(1-x^2)^{\sigma /2}\,d\mu _\lambda (x)\\&\times \,\int _{-1}^{1}u(x)A_\sigma ^{\lambda } u(x)\,d\mu _\lambda (x), \end{aligned}$$

where $Q_{\sigma ,\lambda }$ is the constant given in (5).

Pitt inequality (6) allows us to prove a logarithmic uncertainty principle for the ultraspherical expansions. The main idea comes from [3]. By an elementary argument, for a derivable function such that $\phi (0)=0$ and $\phi (\sigma )>0$ for $\sigma \in (0,\varepsilon )$, with $\varepsilon >0$, it is verified that $\phi '(0_+)\ge 0$. Then, taking the function

$$\begin{aligned} \phi (\sigma )= \sum _{n=0}^\infty \frac{\Gamma (n+\lambda +\frac{1+\sigma }{2})}{\Gamma (n+\lambda +\frac{1-\sigma }{2})} (a_n^\lambda (u))^2 -Q_{\sigma ,\lambda }\int _{-1}^{1}\frac{u^2(x)}{(1-x^2)^{\sigma /2}} \,d\mu _\lambda (x), \end{aligned}$$

we have $\phi (0)=0$ (this is Parseval identity) and, by (6), $\phi (\sigma )>0$ for $\sigma \in (0,1)$, then $\phi '(0_+)\ge 0$ and this inequality gives the logarithmic uncertainty principle, which is written as

$$\begin{aligned} \left( \log 2+\psi \left( \frac{\lambda }{2}+\frac{1}{4}\right) \right)&\int _{-1}^1 u^2(x)\, d\mu _\lambda (x)\\&\le \sum _{n=0}^\infty \psi \left( n+\lambda +\frac{1}{2}\right) (a_n(u))^2\\&\quad +\,\int _{-1}^1\log (\sqrt{1-x^2})u^2(x)\, d\mu _\lambda (x), \end{aligned}$$

where $\psi (a)=\frac{\Gamma '(a)}{\Gamma (a)}$.

In next section we will show an application of Theorem 1 to obtain a Hardy inequality on the sphere. The results in Sect. 3 are the main ingredients in the proof of Theorem 1 which is given in last section of the paper.

2 An Application to the Sphere

It is well known that $L^2(\mathbb {S}^d)=\oplus _{n=0}^\infty \mathcal {H}_n(\mathbb {S}^d)$, where $\mathcal {H}_n(\mathbb {S}^d)$ is the set of spherical harmonics of degree n in $d+1$ variables. If we consider the shifted Laplacian on the sphere

$$\begin{aligned} -\Delta _{\mathbb {S}^d}=\tilde{-\Delta _{\mathbb {S}^d}} +\left( \frac{d-1}{2}\right) ^2, \end{aligned}$$

where $\tilde{-\Delta _{\mathbb {S}^d}}$ is the Laplace-Beltrami operator on $\mathbb {S}^d$, it is verified that

$$\begin{aligned} -\Delta _{\mathbb {S}^d} \mathcal {H}_n(\mathbb {S}^d)=\left( n+\frac{d-1}{2}\right) ^2 \mathcal {H}_n(\mathbb {S}^d). \end{aligned}$$

In this way, the analogous of the operator $A_\sigma ^{\lambda }$ on $\mathbb {S}^d$ is defined by

$$\begin{aligned} \mathbf {A}_\sigma f&=\frac{\Gamma \left( \sqrt{-\Delta _{\mathbb {S}^d}}+\frac{1+\sigma }{2}\right) }{\Gamma \left( \sqrt{-\Delta _{\mathbb {S}^d}}+\frac{1-\sigma }{2}\right) }f\\&=\sum _{n=0}^\infty \frac{\Gamma \left( n+\frac{d-1}{2}+\frac{1+\sigma }{2}\right) }{\Gamma \left( n+\frac{d-1}{2}+\frac{1-\sigma }{2}\right) } {\text {proj}}_{\mathcal {H}_n(\mathbb {S}^d)}f, \end{aligned}$$

where ${\text {proj}}_{\mathcal {H}_n(\mathbb {S}^d)}f$ denotes the projection of f onto the eigenspace $\mathcal {H}_n(\mathbb {S}^d)$.

The operator $\mathbf {A}_\sigma$ becomes the fractional powers of the Laplacian in the Euclidean space through conformal transforms as was observed by Branson in [6]. So $\mathbf {A}_\sigma$ is the natural operator to prove a Hardy type inequality on the sphere. In our proof, we will write $\mathbf {A}_\sigma$ in terms of $A_\sigma ^\lambda$ and this is the main reason to consider $A_\sigma ^\lambda$ in the case of the ultraspherical expansions. An analogous of the Hardy-Littlewood-Sobolev inequality for $\mathbf {A}_\sigma$ and some other inequalities for it were given by Beckner in [2]. The operators $\mathbf {A}_\sigma$ also appear in [18, p. 151] and [17, p. 525].

Each point $x\in \mathbb {S}^d$ can be written as

$$\begin{aligned} x=(t,\sqrt{1-t^2}x'_1,\dots ,\sqrt{1-t^2}x'_d), \end{aligned}$$

for $t\in (-1,1)$ and $x':=(x'_1,\dots ,x'_d)\in \mathbb {S}^{d-1}$, and so

$$\begin{aligned} \int _{\mathbb {S}^d} f(x)\, dx=\int _{-1}^1 \int _{\mathbb {S}^{d-1}}f(t,\sqrt{1-t^2}x')(1-t^2)^{(d-2)/2}\, dx'\, dt. \end{aligned}$$

With these coordinates, see [19, Sect. 3], we have that an orthonormal basis for each $\mathcal {H}_n(\mathbb {S}^d)$ is given by

$$\begin{aligned} \phi _{n,j,k}(x)=\psi _{n,j}(t)Y_{j,k}^d (x'), \qquad j=0,\dots ,n, \end{aligned}$$

with

$$\begin{aligned} \psi _{n,j}(t)=(1-t^2)^{j/2}c_{n-j}^{j+(d-1)/2}(t) \end{aligned}$$

and $\{Y_{j,k}^d\}_{k=1,\dots ,d(j)}$ an orthonormal basis of spherical harmonics on $\mathbb {S}^{d-1}$ of degree j. The value d(j) indicates the dimension of $\mathcal {H}_j(\mathbb {S}^{d-1})$; i.e.,

$$\begin{aligned} d(j)=(2j+d-2)\frac{(j+d-3)!}{j!(d-2)!}. \end{aligned}$$

Then, the orthogonal projection of f onto the eigenspace $\mathcal {H}_n(\mathbb {S}^d)$ can be written as

$$\begin{aligned} {\text {proj}}_{\mathcal {H}_n(\mathbb {S}^d)}f=\sum _{j=0}^n \sum _{k=1}^{d(j)} f_{n,j,k} \phi _{n,j,k}, \end{aligned}$$

with

$$\begin{aligned} f_{n,j,k}=\int _{-1}^{1} G_{j,k}(t) c_{n-j}^{j+(d-1)/2}(t) (1-t^2)^{j+(d-2)/2}\, dt, \end{aligned}$$

$$\begin{aligned} G_{j,k}(t)=(1-t^2)^{-j/2}F_{j,k}(t)\quad \text { and }\quad F_{j,k}(t)=\int _{\mathbb {S}^{d-1}}f(t,\sqrt{1-t^2}x')Y_{j,k}^d(x')\, dx'. \end{aligned}$$

It is easy to observe that

$$\begin{aligned} f(x)=\sum _{j=0}^\infty \sum _{k=1}^{d(j)} F_{j,k}(t)Y_{j,k}^d(x')= \sum _{j=0}^\infty \sum _{k=1}^{d(j)} (1-t^2)^{j/2}G_{j,k}(t)Y_{j,k}^d(x'). \end{aligned}$$

Moreover, from the definition of $\mathbf {A}_\sigma$, we have

$$\begin{aligned} \mathbf {A}_\sigma f(x)=\sum _{j=0}^\infty \sum _{k=1}^{d(j)}(1-t^2)^{j/2} A_{\sigma }^{j+(d-1)/2} G_{j,k}(t)Y_{j,k}^d(x'). \end{aligned}$$

Now, considering the Sobolev space

$$\begin{aligned} \mathbf {H}^{\sigma }=\Big \{f\in L^2(\mathbb {S}^d): \Vert f\Vert _{\mathbf {H}^{\sigma }}:=\Big (\sum _{n=0}^\infty \Big (n+\frac{d-1}{2}\Big )^{\sigma } \Vert {\text {proj}}_{\mathcal {H}_n(\mathbb {S}^d)}f\Vert ^2_{L^2(\mathbb {S}^d)}\Big )^{1/2}<\infty \Big \}, \end{aligned}$$

we have the following Hardy inequality on the sphere.

Theorem 3

Let $d\ge 2$, $0<\sigma <1$, and $e_d$ be the north pole of the sphere $\mathbb {S}^d$. Then for $f\in \mathbf {H}^\sigma$

$$\begin{aligned} 2^\sigma Q_{\sigma ,(d-1)/2}\int _{\mathbb {S}^d}\frac{f^2(x)}{(|x-e_d||x+e_d|)^{\sigma }}\,dx\le \int _{\mathbb {S}^d}f(x)\mathbf {A}_\sigma f(x)\,dx, \end{aligned}$$

(7)

where $Q_{\sigma ,(d-1)/2}$ is the constant given in (5).

Proof

By the orthogonality of the spherical harmonics, it is elementary to show that

$$\begin{aligned} \int _{\mathbb {S}^d}f(x)\mathbf {A}_\sigma f(x)\,dx=\sum _{j=0}^\infty \sum _{k=1}^{d(j)}\int _{-1}^1 G_{j,k}(t) A_{\sigma }^{j+(d-1)/2} G_{j,k}(t)\, d\mu _{j+(d-1)/2}. \end{aligned}$$

Now, applying Theorem 1, we deduce that

$$\begin{aligned} \int _{\mathbb {S}^d}f(x)\mathbf {A}_\sigma f(x)\,dx\ge \sum _{j=0}^\infty \sum _{k=1}^{d(j)}Q_{\sigma , j+(d-1)/2} \int _{-1}^1 \frac{F_{j,k}^2(t)}{(1-t^2)^{\sigma /2}} \, d\mu _{(d-1)/2}. \end{aligned}$$

It is known (see [20]) that for $0<x\le y$ and $j\ge 0$ we have that $\frac{\Gamma (j+y)}{\Gamma (j+x)}\ge \frac{\Gamma (y)}{\Gamma (x)}$. So, $Q_{\sigma ,j+(d-1)/2}\ge Q_{\sigma ,(d-1)/2}$ and

$$\begin{aligned} \int _{\mathbb {S}^d}f(x)\mathbf {A}_\sigma f(x)\,dx\ge Q_{\sigma ,(d-1)/2} \sum _{j=0}^\infty \sum _{k=1}^{d(j)}\int _{-1}^1 \frac{F_{j,k}^2(t)}{(1-t^2)^{\sigma /2}} \, d\mu _{(d-1)/2}. \end{aligned}$$

The proof of (7) is finished by using the identity

$$\begin{aligned} \sum _{j=0}^\infty \sum _{k=1}^{d(j)}\int _{-1}^1 \frac{F_{j,k}^2(t)}{(1-t^2)^{\sigma /2}} \, d\mu _{(d-1)/2}= 2^\sigma \int _{\mathbb {S}^d}\frac{f^2(x)}{(|x-e_d||x+e_d|)^{\sigma }}\, dx. \end{aligned}$$

$\square$

The analogous role on the sphere of radially symmetric functions is played by functions which are invariant under the action of $SO(d-1)$. By $SO(d-1)$-invariance we mean that f is invariant under the action of the group $SO(d-1)$ on $\mathbb {S}^{d-1}$ whenever $SO(d-1)$ is embedded into SO(d) in a suitable way. Each function f of this kind can be written as $f(x)=g(\langle x,e_d\rangle )$, for a certain function g defined in $(-1,1)$. Then for this kind of functions Theorem 3 reduces to Theorem 1 with $\lambda =(d-1)/2$, in this way we can deduce that the constant $2^\sigma Q_{\sigma ,(d-1)/2}$ in (7) is sharp.

As in the classic case, from Theorem 3 we deduce that in a distributional sense

$$\begin{aligned} \mathbf {A}_\sigma -\frac{2^\sigma Q_{\sigma ,(d-1)/2}}{(|x-e_d||x+e_d|)^{\sigma }}\ge 0. \end{aligned}$$

Note that in this case we are perturbing the operator $\mathbf {A}_\sigma$ adding a potential with singularities in both poles of the sphere.

3 Auxiliary Results

The following lemmas give the tools to prove Theorem 1. To be more precise, Lemma 1 provides a nonlocal representation of the operator $A_\sigma ^{\lambda }$ with a kernel having nice properties for our target. Lemma 2 shows the action of the operator $A_\sigma ^{\lambda }$ on the family of weights $(1-x^2)^{-(\lambda /2+(1-\sigma )/4)}$.

For $f,g\in L_\lambda ^2$ we are going to set up the notation

$$\begin{aligned} \langle f,g\rangle _\lambda =\int _{-1}^{1}f(x)g(x)\,d\mu _\lambda (x) \end{aligned}$$

to simplify the writing.

Lemma 1

Let $\lambda >0$ and $0<\sigma <1$. If f is a finite linear combination of ultraspherical polynomials, then

$$\begin{aligned} A_\sigma ^{\lambda } f(x) = \int _{-1}^1\left( f(x)-f(y)\right) K_\sigma ^\lambda (x,y)\,d\mu _\lambda (y) +E_{\sigma ,\lambda }f(x), \qquad x\in (-1,1), \end{aligned}$$

(8)

where the kernel is given by

$$\begin{aligned} K_\sigma ^\lambda (x,y) = D_{\sigma ,\lambda }\int _{-1}^{1}\frac{d\mu _{\lambda -1/2}(t)}{(1-xy-\sqrt{1-x^2}\sqrt{1-y^2}t)^{\lambda +(1+\sigma )/2}}, \end{aligned}$$

with

$$\begin{aligned} D_{\sigma ,\lambda }=\frac{c_\lambda ^2}{2^{\lambda +(1+\sigma )/2}} \frac{\Gamma (\frac{1-\sigma }{2})\Gamma (\lambda +\frac{1+\sigma }{2})}{|\Gamma (-\sigma )|\Gamma (1+\lambda )}, \qquad c_\lambda =\frac{\Gamma (2\lambda +1)}{2^{2\lambda }(\Gamma (\lambda +1/2))^2}, \end{aligned}$$

and

$$\begin{aligned} E_{\sigma ,\lambda }=\frac{\Gamma (\lambda +\frac{1+\sigma }{2})}{\Gamma (\lambda +\frac{1-\sigma }{2})}. \end{aligned}$$

Moreover, for $f\in H_\lambda ^\sigma$ we have

$$\begin{aligned} \langle A_\sigma ^\lambda f, f\rangle _\lambda =\frac{1}{2}\int _{-1}^1\int _{-1}^1 (f(x)-f(y))^2 K_\sigma ^\lambda (x,y) \, d\mu _\lambda (y)\, d \mu _\lambda (x)+E_{\sigma ,\lambda }\langle f, f \rangle _\lambda \end{aligned}$$

(9)

Proof

We start with the identity

$$\begin{aligned} \int _{0}^{\infty }\left( e^{-(n+\lambda )t}-e^{-(\sigma -1)t/2}\right) \left( \sinh t/2\right) ^{-\sigma -1}\,dt = 2^{1+\sigma }\Gamma (-\sigma )\frac{\Gamma (n+\lambda +\frac{1+\sigma }{2})}{\Gamma (n+\lambda +\frac{1-\sigma }{2})} \end{aligned}$$

(10)

for $\lambda >0$ (actually it is also true for values $\lambda >-1/2$) and $0<\sigma <1$. To deduce the previous identity it is enough to apply integration by parts with $u=e^{-(n+\lambda +(1-\sigma )/2)t}-1$ and $v=-2e^{-\sigma t/2}(\sinh t/2)^{-\sigma }/\sigma$, and use [14, Eq. 8, p. 367]

$$\begin{aligned} \int _{0}^{\infty }e^{-\rho t}\left( \cosh (ct)-1\right) ^{\nu }\,dt=\frac{\Gamma (\frac{\rho }{c}-\nu )\Gamma (2\nu +1)}{2^\nu c\Gamma (\frac{\rho }{c}+\nu +1)} \end{aligned}$$

for $c>0$, $2\nu >-1$, and $\rho >c\nu$.

Now, we consider the Poisson operator for ultraspherical expansions. It is given by

$$\begin{aligned} e^{-t\sqrt{-\mathcal {L}_\lambda }}f(x)=\sum _{n=0}^{\infty }e^{-(n+\lambda )t} a_n^\lambda (f)c_n^\lambda (x)=\int _{-1}^1 f(y)P_t^\lambda (x,y)\, d\mu _\lambda (y), \end{aligned}$$

with

$$\begin{aligned} P_t^\lambda (x,y)=\sum _{n=0}^\infty e^{-(n+\lambda )t}c_n^\lambda (x)c_n^\lambda (y). \end{aligned}$$

By the product formula for ultraspherical polynomials [8, Eq. B.2.9, p. 419]

$$\begin{aligned} \frac{C_n^\lambda (x)C_n^\lambda (y)}{C_n^\lambda (1)}=c_\lambda \int _{-1}^{1} C_n^\lambda (xy+\sqrt{1-x^2}\sqrt{1-y^2}t)\,d\mu _{\lambda -1/2}(t), \qquad \lambda >0, \end{aligned}$$

the identity [8, Eq. B.2.8. p. 419]

$$\begin{aligned} \sum _{n=0}^{\infty }\frac{n+\lambda }{\lambda }C_n^\lambda (x)r^n=\frac{1-r^2}{(1-2xr+r^2)^{\lambda +1}}, \qquad 0\le r<1, \end{aligned}$$

and the relation $d_n^2=\frac{\lambda }{c_\lambda (n+\lambda )} C_n^\lambda (1)$, we deduce the expression

$$\begin{aligned} P_t^\lambda (x,y)=\frac{c_\lambda ^2}{2^\lambda } \int _{-1}^{1}\frac{\sinh t}{(\cosh t -w(s))^{\lambda +1}}\,d\mu _{\lambda -1/2}(s), \end{aligned}$$

with $w(s)=xy+\sqrt{1-x^2}\sqrt{1-y^2}s$. The previous identity for $P_t^\lambda$ is not new, it appears as formula (2.12) in [12].

Combining (10) and the definition of the Poisson operator, it is clear that

$$\begin{aligned} A_\sigma ^{\lambda } f(x) = \frac{1}{2^{1+\sigma }\Gamma (-\sigma )}\int _{0}^{\infty } \left( e^{-t\sqrt{-\mathcal {L}_{\lambda }}}f(x)-f(x)e^{-(\sigma -1)t/2} \right) \left( \sinh t/2\right) ^{-\sigma -1}\,dt, \end{aligned}$$

which can be splitted in

$$\begin{aligned}&A_\sigma ^{\lambda } f(x) \nonumber \\&\quad = \frac{1}{2^{1+\sigma }\Gamma (-\sigma )}\int _{0}^{\infty } \left( e^{-t\sqrt{-\mathcal {L}_\lambda }}f(x)-f(x)e^{-t\sqrt{-\mathcal {L}_\lambda }} 1(x)\right) \left( \sinh t/2\right) ^{-\sigma -1}\,dt \nonumber \\&\qquad +\,\frac{f(x)}{2^{1+\sigma }\Gamma (-\sigma )}\int _{0}^{\infty } \left( e^{-t\sqrt{-\mathcal {L}_\lambda }}1(x)-e^{-(\sigma -1)t/2}\right) \left( \sinh t/2\right) ^{-\sigma -1}\,dt. \end{aligned}$$

(11)

From the obvious identity

$$\begin{aligned} e^{-t\sqrt{-\mathcal {L}_\lambda }}1(x)=\int _{-1}^{1}P_t^\lambda (x,y)\, d\mu _\lambda (y)=e^{-\lambda t}, \end{aligned}$$

for the second term in (11) we have

$$\begin{aligned}&\frac{f(x)}{2^{1+\sigma }\Gamma (-\sigma )}\int _{0}^{\infty } \left( e^{-t\sqrt{-\mathcal {L}_\lambda }}1(x)-e^{-(\sigma -1)t/2}\right) \left( \sinh t/2\right) ^{-\sigma -1}\,dt \\&\quad = \frac{f(x)}{2^{1+\sigma }\Gamma (-\sigma )}\int _{0}^{\infty }\left( e^{-\lambda t}-e^{-(\sigma -1)t/2}\right) \left( \sinh t/2\right) ^{-\sigma -1}\,dt\\&\quad = E_{\sigma ,\lambda }f(x), \end{aligned}$$

where we have used (10) with $n=0$.

The first integral in (11) verifies

$$\begin{aligned}&\frac{1}{2^{1+\sigma }\Gamma (-\sigma )}\int _{0}^{\infty } \left( e^{-t\sqrt{-\mathcal {L}_\lambda }}f(x)-f(x)e^{-t\sqrt{-\mathcal {L}_\lambda }} 1(x)\right) \left( \sinh t/2\right) ^{-\sigma -1}\,dt\\&\quad = \frac{1}{2^{1+\sigma }|\Gamma (-\sigma )|}\int _{0}^{\infty }\int _{-1}^{1} P_t^\lambda (x,y)(f(x)-f(y))\, d\mu _\lambda (y)\left( \sinh t/2\right) ^{-\sigma -1}\,dt\\&\quad = \frac{1}{2^{1+\sigma }|\Gamma (-\sigma )|}\int _{-1}^{1}\left( f(x)-f(y)\right) \int _{0}^{\infty }P_t^\lambda (x,y)\left( \sinh t/2\right) ^{-\sigma -1}\,dt\,d\mu _\lambda (y)\\&\quad =\int _{-1}^{1}\left( f(x)-f(y)\right) K_\sigma ^\lambda (x,y)\,d\mu _\lambda (y), \end{aligned}$$

with

$$\begin{aligned} K_\sigma ^\lambda (x,y)= \frac{1}{2^{1+\sigma }|\Gamma (-\sigma )|}\int _{0}^{\infty }P_t^\lambda (x,y)\left( \sinh t/2\right) ^{-\sigma -1}\,dt. \end{aligned}$$

In last computation we have used Fubini theorem. This is justified for finite combinations of ultraspherical polynomials by using the estimate

$$\begin{aligned} P_t^\lambda (x,y)\le \frac{C \sinh t}{(1-x^2)^{\lambda /2}(1-y^2)^{\lambda /2}(\cosh t-xy-\sqrt{1-x^2}\sqrt{1-y^2})}, \end{aligned}$$

which follows from the elementary inequality

$$\begin{aligned} \int _{-1}^1 \frac{(1-s^2)^{\lambda -1}}{(A-Bs)^{\lambda +1}}\, ds\le \frac{C}{B^\lambda (A-B)}, \qquad A>B>0, \quad \lambda >0, \end{aligned}$$

and the mean value theorem.

Indeed, taking $C_f=\max \{|f'(x)|:x\in [-1,1]\}$ and using the inequality $1-xy-\sqrt{1-x^2}\sqrt{1-y^2}\ge C|x-y|^2$, we have

$$\begin{aligned}&\int _{0}^{\infty }\int _{-1}^{1} P_t^\lambda (x,y)|f(x)-f(y)|\, d\mu _\lambda (y)\left( \sinh t/2\right) ^{-\sigma -1}\,dt\\&\quad \le \frac{C_f}{(1-x^2)^{\lambda /2}} \left( C_1\int _0^1\int _{-1}^1 \frac{t^{-\sigma }|x-y|}{t^2+|x-y|^2}(1-y^2)^{\lambda /2-1/2}\, dy \, dt\right. \\&\qquad \left. +\,C_2\int _1^\infty \int _{-1}^1 e^{-(\sigma +1)t/2}|x{-}y|(1-y^2)^{\lambda /2-1/2}\, dy \, dt\right) =: \frac{C_f}{(1-x^2)^{\lambda /2}} ( I_1+I_2). \end{aligned}$$

Obviously, $I_2$ is a finite integral. For $I_1$ the change of variable $t=|x-y|s$ gives

$$\begin{aligned} I_1\le C_1 \int _0^\infty \frac{s^{-\sigma }}{s^2+1}\, ds\int _{-1}^1 |x-y|^{-\sigma }(1-y^2)^{\lambda /2-1/2}\, dy<\infty . \end{aligned}$$

To obtain the expression of $K_\sigma ^\lambda$ we observe that

$$\begin{aligned}&K_\sigma ^\lambda (x,y)\\&\quad =\frac{c_\lambda ^2}{2^{\lambda +1+\sigma }|\Gamma (-\sigma )|} \int _{0}^{\infty }\int _{-1}^{1}\frac{\sinh t}{(\cosh t -w(s))^{\lambda +1}}\,d\mu _{\lambda -1/2}(s)\left( \sinh t/2\right) ^{-\sigma -1}\,dt\\&\quad = \frac{c_\lambda ^2}{2^{\lambda +(1+\sigma )/2}} \frac{\Gamma (\frac{1-\sigma }{2})\Gamma (\lambda +\frac{1+\sigma }{2})}{|\Gamma (-\sigma )|\Gamma (\lambda +1)} \int _{-1}^{1}\frac{d\mu _{\lambda -1/2}(s)}{(1-w(s))^{\lambda +(1+\sigma )/2}}, \end{aligned}$$

where we have applied Fubini theorem and the change of variable $2(\sinh t/2)^2=z(1-w(s))$ in last equality. With the last identity we have concluded the proof of (8).

To prove (9) we follow the argument in [16, Lemma 5.1]. First, we observe that the kernel $K_\sigma ^\lambda (x,y)$ is positive and symmetric in the sense that $K_\sigma ^\lambda (x,y)=K_\sigma ^\lambda (y,x)$. Then, (9) is clear when f is a finite linear combination of ultraspherical polynomials. For $f\in H_\lambda ^\sigma$ we consider a sequence of finite linear combinations of ultraspherical polynomials $\{p_k\}_{k\ge 0}$ such that $p_k$ converges to f in $H_\lambda ^\sigma$. Then, by using the definition of $A_\sigma ^\lambda$, it is clear that $\langle A_\sigma ^\lambda p_k, p_k\rangle _{\lambda }$ converges to $\langle A_\sigma ^\lambda f, f\rangle _{\lambda }$. Moreover, the result for polynomial functions implies

$$\begin{aligned} \langle A_\sigma ^\lambda p_k, p_k\rangle _{\lambda }= & {} \frac{1}{2}\int _{-1}^1\int _{-1}^1 (p_k(x)-p_k(y))^2 K_\sigma ^\lambda (x,y) \, d\mu _\lambda (y)\, d \mu _\lambda (x)\nonumber \\&+\,E_{\sigma ,\lambda }\langle p_k, p_k \rangle _\lambda <\infty . \end{aligned}$$

(12)

Consequently, the functions $P_k(x,y)=p_k(x)-p_k(y)$ form a Cauchy sequence in $L^2((-1,1)\times (-1,1), d\omega )$ where $d\omega (x,y)=K_\sigma ^\lambda (x,y)\, d\mu _\lambda (x)\, d\mu _\lambda (y)$ which converges to $f(x)-f(y)$ in this norm. Hence, passing to the limit in (12), we complete the proof of the lemma. $\square$

Lemma 2

Let $\lambda >0$ and $2\lambda +1>\sigma >0$. Then

$$\begin{aligned} A_\sigma ^{\lambda }\left( \frac{1}{(1-x^2)^{\lambda /2+(1-\sigma )/4}}\right) = \frac{Q_{\sigma ,\lambda }}{(1-x^2)^{\lambda /2+(1+\sigma )/4}}, \end{aligned}$$

(13)

where $Q_{\sigma ,\lambda }$ is the constant given in (5).

Proof

First of all, we have to realize that the ultraspherical polynomial $C_{n}^\lambda (x)$ is odd for $n=2m+1$, $m\in \mathbb {Z}^{+}$; therefore, for $\beta >0$, the function $(1-x^2)^{\beta -1} C_{2m+1}^\lambda (x)$ is an odd function and its integral over the interval $(-1,1)$ is zero. For $n=2m$ we use [15, Eq. 15, p. 519] to obtain

$$\begin{aligned}&\int _{-1}^{1}(1-x^2)^{\beta -1}C_{2m}^\lambda (x)\,dx \\&\quad = \sqrt{\pi }\frac{(2\lambda )_{2m}}{(2m)!} \frac{\Gamma (\beta )}{\Gamma (\beta +1/2)} {}_3F_2(-2m,2\lambda +2m,\beta ;2\beta ,\lambda +1/2;1)\\&\quad = \pi \frac{(2\lambda )_{2m}}{(2m)!} \frac{\Gamma (\beta )\Gamma (\lambda +1/2)\Gamma (\beta -\lambda +1/2)}{\Gamma (1/2-m)\Gamma (\lambda +m+1/2)\Gamma (\beta +m+1/2) \Gamma (\beta -\lambda -m+1/2)}, \end{aligned}$$

where in last identity we have evaluated the hypergeometric function with the so-called Watson formula [13, Eq. 16.4.6, p. 406]. Therefore, if we denote $\alpha =\lambda /2+(1-\sigma )/4$, we obtain that

$$\begin{aligned} \int _{-1}^{1}(1-x^2)^{\alpha -1}C_{2m}^\lambda (x)\,dx= R_{\sigma ,\lambda }\int _{-1}^{1}(1-x^2)^{\alpha +\sigma /2-1} C_{2m}^{\lambda }(x)\,dx, \end{aligned}$$

(14)

with

$$\begin{aligned} R_{\sigma ,\lambda }= & {} \frac{\Gamma (\alpha )\Gamma (\alpha -\lambda +1/2)}{\Gamma (\alpha +\sigma /2)\Gamma (\alpha -\lambda +1/2+\sigma /2)}\\&\times \,\frac{\Gamma (\alpha +m+1/2+\sigma /2)\Gamma (\alpha -\lambda -m+1/2+\sigma /2)}{ \Gamma (\alpha +m+1/2)\Gamma (\alpha -\lambda -m+1/2)}. \end{aligned}$$

In this way, if we prove the identity

$$\begin{aligned} R_{\sigma ,\lambda }=Q_{\sigma ,\lambda }^{-1}\frac{\Gamma (2m+2\alpha +\sigma )}{\Gamma (2m+2\alpha )} \end{aligned}$$

(15)

we will conclude the proof, because (14) implies

$$\begin{aligned} a_n^\lambda \left( \frac{1}{(1-x^2)^{\alpha +\sigma /2}}\right) =Q_{\sigma ,\lambda }^{-1}\frac{\Gamma (n+2\alpha +\sigma )}{\Gamma (n+2\alpha )} a_n^\lambda \left( \frac{1}{(1-x^2)^\alpha }\right) , \end{aligned}$$

where we have had in mind that the n-th Fourier coefficient is null when $n=2m+1$.

Let us check that (15) actually holds. Using the reflection formula [1, Eq. 6.1.17, p. 256] twice we have

$$\begin{aligned} \frac{\Gamma (\alpha -\lambda -m+1/2+\sigma /2)}{\Gamma (\alpha -\lambda -m+1/2)}&= \frac{\Gamma (\alpha +m+\sigma /2)}{\Gamma (\alpha +m)} \frac{\sin (\pi (\alpha -\lambda -m+1/2))}{\sin (\pi (\alpha -\lambda -m+1/2+\sigma /2))} \\&= \frac{\Gamma (\alpha +m+\sigma /2)}{\Gamma (\alpha +m)} \frac{\Gamma (\alpha )\Gamma (\alpha -\lambda +1/2+\sigma /2)}{\Gamma (\alpha +\sigma /2)\Gamma (\alpha -\lambda +1/2)}, \end{aligned}$$

and then

$$\begin{aligned} R_{\sigma ,\lambda }&= \frac{\Gamma (\alpha )^2}{\Gamma (\alpha +\sigma /2)^2} \frac{\Gamma (\alpha +m+\sigma /2)\Gamma (\alpha +m+\sigma /2+1/2)}{\Gamma (\alpha +m)\Gamma (\alpha +m+1/2)}\\&= Q_{\sigma ,\lambda }^{-1}\frac{\Gamma (2m+2\alpha +\sigma )}{\Gamma (2m+2\alpha )}, \end{aligned}$$

by the duplication formula [1, Eq. 6.1.18, p. 256]. $\square$

4 Proof of Theorem 1

Polarizing the identity (9) in Lemma 1 we obtain

$$\begin{aligned} \langle g,A_\sigma ^{\lambda } f\rangle _\lambda =\frac{1}{2}\int _{-1}^{1}\int _{-1}^{1}F(x,y) K_\sigma ^\lambda (x,y)\,d\mu _\lambda (y)\,d\mu _\lambda (x)+E_{\sigma ,\lambda }\langle g,f\rangle _\lambda , \end{aligned}$$

(16)

with $F(x,y)=(g(x)-g(y))(f(x)-f(y))$.

Let us take $g(x)=(1-x^2)^{-\lambda /2-(1-\sigma )/4}$ and $f(x)=u^2(x)/g(x)$ for $u\in H_\lambda ^\sigma$. Then

$$\begin{aligned} F(x,y)=\left( u(x)-u(y)\right) ^2-g(x)g(y)\left( \frac{u(x)}{g(x)} -\frac{u(y)}{g(y)}\right) ^2 \end{aligned}$$

and (16) becomes

$$\begin{aligned}&\langle g,A_\sigma ^{\lambda } f\rangle _\lambda \\&\quad = \langle u,A_\sigma ^{\lambda } u\rangle _\lambda -\frac{1}{2}\int _{-1}^{1}\int _{-1}^{1}g(x)g(y) \left( \frac{u(x)}{g(x)}-\frac{u(y)}{g(y)}\right) ^2 K_\sigma ^\lambda (x,y)\,d\mu _\lambda (y)\,d\mu _\lambda (x). \end{aligned}$$

Now, by (13), we have

$$\begin{aligned} \langle g,A_\sigma ^{\lambda } f\rangle _\lambda = \langle A_\sigma ^{\lambda } g,f\rangle _\lambda = Q_{\sigma ,\lambda }\int _{-1}^{1}\frac{u^2(x)}{(1-x^2)^{\sigma /2}}\,d\mu _\lambda (x) \end{aligned}$$

and then we can deduce the ground state representation

$$\begin{aligned} \langle u,A_\sigma ^{\lambda } u\rangle _\lambda -Q_{\sigma ,\lambda }&\int _{-1}^{1}\frac{u^2(x)}{(1-x^2)^{\sigma /2}}\,d\mu _\lambda (x) \nonumber \\= & {} \frac{1}{2}\int _{-1}^{1}\int _{-1}^{1}g(x)g(y) \left( \frac{u(x)}{g(x)}-\frac{u(y)}{g(y)}\right) ^2\nonumber \\&\times \,K_\sigma ^\lambda (x,y)\,d\mu _\lambda (y)\,d\mu _\lambda (x). \end{aligned}$$

(17)

So, due to the positivity of the kernel $K_{\sigma }^\lambda$, we conclude the proof.