A Hardy inequality for ultraspherical expansions with an application to the sphere

TL;DR

This paper establishes a Hardy inequality for ultraspherical expansions using ground state methods, leading to uncertainty principles and a Hardy inequality on spheres with double singularities.

Contribution

It introduces a Hardy inequality for ultraspherical expansions and applies it to derive uncertainty principles and sphere inequalities with singular potentials.

Findings

Hardy inequality for ultraspherical expansions proved

Uncertainty principles derived from the inequality

Hardy inequality on spheres with double singularity

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.

Equations

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

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

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

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

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

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

$$f⟼\sum _{n=0}^{\infty }{a}_n^{\lambda }(f)c_n^{\lambda },$$

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

$$(-{\mathcal{L}}_{\lambda }{)}^{\sigma /2}f=\sum _{n=0}^{\infty }(n+\lambda {)}^{\sigma }{a}_n^{\lambda }(f)c_n^{\lambda },\sigma >0.$$

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

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

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

$$H^{\sigma}_{\lambda}=\Big{\{}f\in L^{2}_{\lambda}:\|f\|_{H^{\sigma}_{\lambda}}:=\Big{(}\sum_{n=0}^{\infty}(n+\lambda)^{\sigma}(a_{n}^{\lambda}(f))^{2}\Big{)}^{1/2}<\infty\Big{\}}.$$

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

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

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

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

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

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

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

$$-{\Delta }_{{\mathbb{S}}^d}=\stackrel{~}{-{\Delta }_{{\mathbb{S}}^d}}+{\left(\frac{d-1}{2}\right)}^2,$$

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

$${\mathbf{A}}_{\sigma }f$$

$$=\sum _{n=0}^{\infty }\frac{\Gamma \left(n+\frac{d-1}{2}+\frac{1+\sigma }{2}\right)}{\Gamma \left(n+\frac{d-1}{2}+\frac{1-\sigma }{2}\right)}{\mathrm{proj}}_{{\mathcal{H}}_n({\mathbb{S}}^d)}f,$$

$$x=(t,\sqrt{1-t^2}{x}_1^{\prime },\dots ,\sqrt{1-t^2}{x}_d^{\prime }),$$

$${\int }_{{\mathbb{S}}^d}f(x)dx={\int }_{-1}^1{\int }_{{\mathbb{S}}^{d-1}}f(t,\sqrt{1-t^2}{x}^{\prime })(1-t^2{)}^{(d-2)/2}d{x}^{\prime }dt.$$

$${\varphi }_{n,j,k}(x)={\psi }_{n,j}(t)Y_{j,k}^d(x^{\prime }),j=0,\dots ,n,$$

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

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

$${\mathrm{proj}}_{{\mathcal{H}}_n({\mathbb{S}}^d)}f=\sum _{j=0}^n\sum _{k=1}^{d(j)}{f}_{n,j,k}{\varphi }_{n,j,k},$$

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

Full text

Research of the second author supported by grant MTM2015-65888-C4-4-P of the Spanish government

1. Introduction and main result

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

[TABLE]

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

[TABLE]

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

[TABLE]

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

[TABLE]

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

[TABLE]

The sequence of polynomials $\{c_{n}^{\lambda}\}_{n\geq 0}$ 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

[TABLE]

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

[TABLE]

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

[TABLE]

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

[TABLE]

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 other 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

[TABLE]

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

[TABLE]

Note that

[TABLE]

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

[TABLE]

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 $u\in H_{\lambda}^{\sigma}$

[TABLE]

where

[TABLE]

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

[TABLE]

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),

[TABLE]

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 $u\in H_{\lambda}^{\sigma}$

[TABLE]

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 $\phi^{\prime}(0_{+})\geq 0$. Then, taking the function

[TABLE]

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

[TABLE]

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

In next section we will show an application of Theorem 1 to obtain a Hardy inequality on the sphere. The results in Section 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

[TABLE]

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

[TABLE]

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

[TABLE]

where $\operatorname{proj}_{\mathcal{H}_{n}(\mathbb{S}^{d})}f$ 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 T. P. 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 W. 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

[TABLE]

for $t\in(-1,1)$ and $x^{\prime}:=(x^{\prime}_{1},\dots,x^{\prime}_{d})\in\mathbb{S}^{d-1}$, and so

[TABLE]

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

[TABLE]

with

[TABLE]

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.,

[TABLE]

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

[TABLE]

with

[TABLE]

[TABLE]

It is easy to observe that

[TABLE]

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

[TABLE]

Now, considering the Sobolev space

[TABLE]

we have the following Hardy inequality on the sphere.

Theorem 3**.**

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

[TABLE]

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

[TABLE]

Now, applying Theorem 1, we deduce that

[TABLE]

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

[TABLE]

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

[TABLE]

∎

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

[TABLE]

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

[TABLE]

to simplify the writing.

Lemma 1**.**

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

[TABLE]

where the kernel is given by

[TABLE]

with

[TABLE]

and

[TABLE]

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

[TABLE]

Proof.

We start with the identity

[TABLE]

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]

[TABLE]

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

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

[TABLE]

with

[TABLE]

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

[TABLE]

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

[TABLE]

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

[TABLE]

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

[TABLE]

which can be splitted in

[TABLE]

From the obvious identity

[TABLE]

for the second term in (11) we have

[TABLE]

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

The first integral in (11) verifies

[TABLE]

with

[TABLE]

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

[TABLE]

which follows from the elementary inequality

[TABLE]

and the mean value theorem. Indeed, taking $C_{f}=\max\{|f^{\prime}(x)|:x\in[-1,1]\}$ and using the inequality $1-xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}}\geq C|x-y|^{2}$, we have

[TABLE]

Obviously, $I_{2}$ is a finite integral. For $I_{1}$ the change of variable $t=|x-y|s$ gives

[TABLE]

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

[TABLE]

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 $\{p_{k}\}_{k\geq 0}$ 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

[TABLE]

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. ∎

Lemma 2**.**

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

[TABLE]

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

[TABLE]

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

[TABLE]

with

[TABLE]

In this way, if we prove the identity

[TABLE]

we will conclude the proof, because (14) implies

[TABLE]

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

[TABLE]

and then

[TABLE]

by the duplication formula [1, eq. 6.1.18, p. 256]. ∎

4. Proof of Theorem 1

Polarizing the identity (9) in Lemma 1 we obtain

[TABLE]

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

[TABLE]

and (16) becomes

[TABLE]

Now, by (13), we have

[TABLE]

and then we can deduce the ground state representation

[TABLE]

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