Iterated convolution inequalities on and Riemannian symmetric spaces of non-compact type

Utsav Dewan

doi:10.4153/S0008414X26102442

Iterated convolution inequalities on $\mathbb {R}^d$ and Riemannian symmetric spaces of non-compact type

Part of: Abstract harmonic analysis Miscellaneous topics - Partial differential equations Lie groups

Published online by Cambridge University Press: 21 September 2026

Utsav Dewan

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Utsav Dewan*: Affiliation:
Department of Mathematics, Indian Institute of Technology Bombay , Powai, Mumbai, India
*: e-mail: utsav@math.iitb.ac.in

Article contents

Abstract
Introduction
Preliminaries
Proofs of Theorem and the Euclidean result
Results on symmetric spaces
Footnotes
References

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Abstract

Let $\mathbb {X}$ bea Riemannian symmetric space of non-compact type. For a real-valued $f \in L^1(\mathbb {X})$ with $\int _{\mathbb {X}} f \ge 0$, we prove that if f satisfies the iterated convolution inequality:

$$ \begin{align*} f \ge \sum_{n=2}^N a_n \left(*^n f\right),\:\:\text{ a.e. on } \mathbb{X}, \end{align*} $$

where $N \ge 2$

is an integer and for $2 \le n \le N$

, $a_n$

are nonnegative integers with at least one of them positive, then f must be nonnegative a.e. and satisfy the nontrivial bound $\int _{\mathbb {X}} f \le t_{\mathcal {Q}}\:$

, where $\mathcal {Q}(t):=t-\sum _{n=2}^N a_n\:t^n$

and $t_{\mathcal {Q}}$

is the unique zero of $\mathcal {Q}'$

in $(0,\infty )$

. This result is new even for $\mathbb {R}^d$

and generalizes recent results of Carlen et al. (2021, Int. Math. Res. Not., 24, 18604–18612) and Nakamura and Sawano (2025, J. Geom. Anal., 35, 68) for two-fold and m-fold convolutions, respectively, to genuine polynomials. We then apply our result to obtain an a priori estimate for solutions of an integro-differential equation on symmetric spaces, which generalizes the equation on $\mathbb R^d$

, related to the physical problem of the ground state energy of the Bose gas, studied by Carlen et al. (2020, Pure Appl. Anal., 2, 659–684). We also obtain a surprising result on the nature of the extremizers of the convolution inequalities.

Keywords

Convolution Riemannian symmetric spaces of non-compact type integro-differential equation

MSC classification

Primary: 22E30: Analysis on real and complex Lie groups 43A85: Analysis on homogeneous spaces

Secondary: 35R09: Integro-partial differential equations

Information

Type: Article
Information: Canadian Journal of Mathematics , First View , pp. 1 - 23

DOI: https://doi.org/10.4153/S0008414X26102442 [Opens in a new window]
Copyright: © The Author(s), 2026. Published by Cambridge University Press on behalf of Canadian Mathematical Society

1 Introduction

While conducting a recent study on the physical problem of the ground state energy of the Bose gas in [Reference Carlen, Jauslin and LiebCJL20], Carlen et al. were propelled to look at the real, integrable solutions of the following convolution inequality almost everywhere on $\mathbb R^d$ :

(1.1) $$ \begin{align} f \ge f * f. \end{align} $$

While it is easy to observe that the equality case of (1.1) fails to produce any nontrivial integrable solution, in [Reference Carlen, Jauslin, Lieb and LossCJLL21], Carlen et al. produced a family of nontrivial radial solutions in $L^1(\mathbb R^d)$ for the inequality (1.1) by considering

$$ \begin{align*} f_{a,t} = \widehat{g_{a,t}}, \end{align*} $$

the Fourier transform of the functions

$$ \begin{align*} g_{a,t}(x):=ae^{-2\pi t\|x\|}, \text{ for } 0 < a \le 1/2\:,\: t>0. \end{align*} $$

They observed that the functions $f_{a,t}$ are nonnegative and satisfy

$$ \begin{align*} \int_{\mathbb R^d} f_{a,t} \le \frac{1}{2}. \end{align*} $$

Then the authors proceeded to show that the above examples $f_{a,t}$ are surprisingly typical of all integrable solutions by proving [Reference Carlen, Jauslin, Lieb and LossCJLL21, Theorem 1]: any real $f \in L^1(\mathbb R^d)$ satisfying the convolution inequality (1.1) is nonnegative with

(1.2) $$ \begin{align} \int_{\mathbb R^d} f \le \frac{1}{2}. \end{align} $$

We note that by simply integrating out the inequality (1.1), one only gets the a priori bound

$$ \begin{align*} \int_{\mathbb R^d} f \le 1, \end{align*} $$

and thus (1.2) is nontrivial. The nonnegativity of f is also special to the case $L^1(\mathbb R)$ as for any $p>1$ , one can get counter-examples belonging to $L^p(\mathbb R)$ by simply considering the Fourier transform of the indicator function of symmetric intervals centered at the origin.

Recently, Nakamura and Sawano generalized the above result of Carlen et al. by studying the integrable solutions of the m-fold convolution inequality on $\mathbb R^d$ for $m \ge 2$ :

(1.3) $$ \begin{align} f \ge \underbrace{f * \cdots *f }_{m-times}. \end{align} $$ f greater than or equals ModifyingBelow f asterisk midline horizontal ellipsis asterisk f With bottom brace Underscript m minus t i m e s Endscripts period

Under the additional condition that $\int _{\mathbb R^d} f \ge 0$ , if m is odd, they showed that [Reference Nakamura and SawanoNS25, Theorems 1.2 and 1.4]: any real $f \in L^1(\mathbb R^d)$ satisfying the convolution inequality (1.3) is nonnegative with

$$ \begin{align*} \int_{\mathbb R^d} f \le m^{-\frac{1}{m-1}}. \end{align*} $$

An inquisitive mind is naturally led to inquire what happens when the iterated convolution inequalities (1.1) or (1.3) are replaced by a genuine polynomial:

(1.4) $$ \begin{align} f \ge \sum_{n=2}^N a_n \left(*^n f\right), \end{align} $$ f greater than or equals sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline left parenthesis asterisk Superscript n Baseline f right parenthesis comma

where $N \ge 2$ is an integer, for $2 \le n \le N$ , $a_n$ are nonnegative integers with at least one of them positive and

$$ \begin{align*} *^n f = \underbrace{f * \cdots *f }_{n-times}. \end{align*} $$ asterisk Superscript n Baseline f equals ModifyingBelow f asterisk midline horizontal ellipsis asterisk f With bottom brace Underscript n minus t i m e s Endscripts period

In this article, we study the convolution inequality (1.4) on $\mathbb R^d$ and Riemannian symmetric spaces of non-compact type. Intimately connected to our analysis of the inequality (1.4) will be the polynomial

(1.5) $$ \begin{align} \mathcal Q(t)=t-\sum_{n=2}^N a_n\:t^n, \end{align} $$ script upper Q left parenthesis t right parenthesis equals t minus sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline t Superscript n Baseline comma

where $a_n$ are as in (1.4). It is easy to observe that $\mathcal Q$ attains a unique positive maximum on $(0,\infty )$ say at $t_{\mathcal Q}$ and

(1.6) $$ \begin{align} t_{\mathcal Q} \in (0,1)\:\:\text{ with }\:\mathcal Q\left(t_{\mathcal Q}\right)<t_{\mathcal Q}. \end{align} $$

Moreover, by Descartes’ rule of signs, the polynomial $\mathcal Q'$ has a unique zero in $(0,\infty )$ and thus is given by $t_{\mathcal Q}$ .

Upon a brief glance at the inequality (1.4), one may inquire whether the class of real-valued $f \in L^1(\mathbb R^d)$ satisfying the iterated convolution inequality (1.4) to begin with is sufficiently rich. Our first result (Theorem 1.1) shows that it is indeed the case for G a second countable, locally compact, Hausdorff topological group and in fact, one can have a concrete realization of the structure of such solutions. In the following statement, $L^1(G)$ is considered with respect to a fixed left Haar measure.

Theorem 1.1 Let $N \ge 2$ be an integer, for $2 \le n \le N$ , $a_n$ be nonnegative integers with at least one of them positive, $\mathcal Q$ be as in (1.5) and $t_{\mathcal Q}$ be the unique zero of $\mathcal Q'$ in $(0,\infty )$ . Let G be a second countable, locally compact, Hausdorff topological group. Now consider any nonnegative $\psi \in L^1(G)$ with

$$ \begin{align*} \|\psi\|_{L^1(G)} \le \mathcal Q\left(t_{\mathcal Q}\right). \end{align*} $$

Define $\Psi _j$ , $j \in \mathbb N \cup \{0\}$ , inductively by

$$ \begin{align*} &\Psi_0 := \psi\:, \\ &\Psi_{j+1} := \psi \:+\:\sum_{n=2}^N a_n \left(*^n \Psi_j\right) \:,\:j \in \mathbb N \cup \{0\}. \end{align*} $$

StartLayout 1st Row 1st Column Blank 2nd Column normal upper Psi 0 colon equals psi comma 2nd Row 1st Column Blank 2nd Column normal upper Psi Subscript j plus 1 Baseline colon equals psi plus sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline left parenthesis asterisk Superscript n Baseline normal upper Psi Subscript j Baseline right parenthesis comma j element of double struck upper N union StartSet 0 EndSet period EndLayout

Then $\{\Psi _j\}_{j=0}^\infty $ converges to an element $\Psi $ in the topology of $L^1(G)$ , which satisfies

$$ \begin{align*} \Psi \ge 0\:,\: \|\Psi\|_{L^1(G)} \le t_{\mathcal Q}\text{ and }\psi = \Psi - \sum_{n=2}^N a_n \left(*^n \Psi\right). \end{align*} $$

Next, we study the properties of the solutions of the inequality (1.4).

Theorem 1.2 Let $N,\:a_n,\:\mathcal Q,\:t_{\mathcal Q}$ be as in Theorem 1.1. Suppose a real-valued $f \in L^1(\mathbb R^d)$ with $\int _{\mathbb R^d} f \ge 0$ satisfies

(1.7) $$ \begin{align} f \ge \sum_{n=2}^N a_n \left(*^n f\right). \end{align} $$ f greater than or equals sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline left parenthesis asterisk Superscript n Baseline f right parenthesis period

Then f is nonnegative and satisfies $\|f\|_{L^1(\mathbb R^d)}\le t_{\mathcal Q}$ .

Remark 1.3

(i) The proof of Theorem 1.1 consists of a combinatorial argument.
(ii) The proof of Theorem 1.2 involves Fourier Analysis and crucial applications of a nontrivial result from complex analysis regarding singularities of power series, known as the Vivanti–Pringsheim theorem (Lemma 2.1).
(iii) Theorem 1.2 is false for the unit circle $\mathbb {T}$ . This is illustrated by a counter-example in Remark 3.1.

We now come to the main results of this article, on Riemannian symmetric spaces of non-compact type. These are homogeneous spaces $\mathbb X=G/K$ , where G is a connected, non-compact, semi-simple Lie group with finite center and K is a maximal compact subgroup of G. All integrals or Lebesgue spaces considered on G will be with respect to a fixed Haar measure, which we denote by $dx$ .

Theorem 1.4 Let $N,\:a_n,\:\mathcal Q,\:t_{\mathcal Q}$ be as in Theorem 1.1. Suppose a real-valued $f \in L^1(\mathbb X)$ , that is, a right K-invariant, real-valued $f \in L^1(G)$ with $\int _G f \ge 0$ , satisfies

(1.8) $$ \begin{align} f \ge \sum_{n=2}^N a_n \left(*^n f\right). \end{align} $$ f greater than or equals sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline left parenthesis asterisk Superscript n Baseline f right parenthesis period

Then f is nonnegative and satisfies $\|f\|_{L^1(G)}\le t_{\mathcal Q}$ .

In the simplest case, when $\mathcal Q(t)=t-t^2$ , we have $t_{\mathcal Q}=1/2$ . Thus, the $L^1$ functions satisfying $f \ge f * f$ almost everywhere, with

$$ \begin{align*} \int_G f= \frac{1}{2}, \end{align*} $$

can be viewed as the extremizers of the convolution inequality. Our next result entails a surprising nature of such K-bi-invariant extremizers.

Theorem 1.5 Let $f \in L^1(G//K)$ such that $f \ge f*f$ and $\int _G f= 1/2$ . Then

$$ \begin{align*} \int_G \|x\|\:f(x)\:dx= \infty. \end{align*} $$

In particular, f cannot be an $L^1$ -Schwartz class function on G.

In this article, $L^1(G//K)$ is the class of K-bi-invariant members of $L^1(G)$ and for $x \in G$ , $\|x\|:=|x^+|$ , where $x^+ \in \overline {\mathfrak {a}^+}$ , is the corresponding component in the Cartan decomposition.

Remark 1.6

(i) Proof of Theorem 1.4 comprises technical subtleties in the form of the operator-valued group Fourier transform, arising from the representation theory of non-compact semi-simple Lie groups. It also crucially utilizes the decay of the ground spherical function away from the identity.
(ii) Theorem 1.5 is obtained by means of a transference method based on the Abel transform.

As an application of Theorem 1.4, we look at an integro-differential equation, which generalizes the equation on $\mathbb R^d$ , related to the physical problem of the ground state energy of the Bose gas, studied by Carlen et al. [Reference Carlen, Jauslin and LiebCJL20]. To state our result, we need to introduce some notations. Let $\Delta $ be the Laplace–Beltrami operator corresponding to the left-invariant Riemannian metric on $\mathbb X$ . The $L^2$ -spectrum of $\Delta $ is given by $(-\infty ,-\|\rho \|^2]$ (for unexplained notations, we refer to Section 2). We now consider the shifted Laplace–Beltrami operator $\mathcal L:= \Delta + \|\rho \|^2$ and the integro-differential equation of a K-bi-invariant function u:

(1.9) $$ \begin{align} {\left(-\mathcal L +\xi \right)}^m u(x) = V(x)(1-u(x))+\mu \left(*^{m+1} u\right)(x)\:,\: x \in G, \end{align} $$

under a constraint

(1.10) $$ \begin{align} \int_G u = \frac{1}{\delta}, \end{align} $$

where m is a positive integer, $\mu \ge 0$ , $\xi ,\delta>0$ are given parameters, and V is a given nonnegative potential in $L^1(G//K)$ . Then we have the a priori estimate.

Theorem 1.7 Let $m \in \mathbb N,\: \mu \ge 0\:,\:\xi , \delta>0$ satisfy

(1.11) $$ \begin{align} \frac{\mu^{1/m}}{\xi\delta} <1. \end{align} $$ StartFraction mu Superscript 1 divided by m Baseline Over xi delta EndFraction less than 1 period

Suppose $V \in L^1(G//K)$ is nonnegative and real $u \in L^1(G//K)$ is a solution of (1.9) and (1.10) such that $u \le 1$ almost everywhere on G. Then u is nonnegative almost everywhere on G.

This article is organized as follows. In Section 2, we recall the necessary preliminaries about Riemannian symmetric spaces of non-compact type and the Fourier analysis thereon. In Section 3, we prove Theorems 1.1 and 1.2. In Section 4, we prove Theorems 1.4, 1.5, and 1.7.

2 Preliminaries

In this section, we recall the essential preliminaries and fix our notations. We refer the reader to [Reference Gangolli and VaradarajanGV88, Reference HelgasonHel08] for more details.

Two nonnegative functions $f_1$ and $f_2$ are defined to satisfy $f_1 \lesssim f_2$ if there exists a constant $C>0$ such that $f_1 \le Cf_2$ . $\mathbb N$ will denote the set of positive integers.

Let G be a connected, non-compact, semi-simple Lie group with finite center, and let K be a maximal compact subgroup of G. Then $\mathbb X=G/K$ is a Riemannian symmetric space of non-compact type. The class of K-bi-invariant functions on G is naturally identified with left K-invariant functions on $\mathbb X$ .

Let $\mathfrak {g}$ be the Lie algebra of G, and let $\mathfrak {g}=\mathfrak {k}+\mathfrak {p}$ be the Cartan decomposition on the Lie algebra level. The Killing form of $\mathfrak {g}$ induces a K-invariant inner product $\langle \cdot ,\cdot \rangle $ on $\mathfrak {p}$ and hence a G-invariant Riemannian metric on $\mathbb X$ .

We fix a maximal abelian subspace $\mathfrak {a}$ of $\mathfrak {p}$ . The rank of $\mathbb X$ is the real rank of G, which is given by the dimension of $\mathfrak {a}$ . $\mathfrak {a}^*$ , the real dual of $\mathfrak {a}$ will be identified with $\mathfrak {a}$ via the inner product inherited from $\mathfrak {p}$ . Let $\Sigma $ denote the set of restricted roots of the pair $(\mathfrak {g}, \mathfrak {a})$ and W be the Weyl group associated with $\Sigma $ . For $\alpha \in \Sigma $ , let $\mathfrak {g}_\alpha $ and $m_\alpha $ denote the root space corresponding to the root $\alpha $ and the dimension of $\mathfrak {g}_\alpha $ , respectively. We make the choice of a positive Weyl chamber $\mathfrak {a}^+ \subset \mathfrak {a}$ , and let $\Sigma ^+$ and $\Sigma ^+_r$ be the corresponding set of positive roots and positive reduced (indivisible) roots, respectively. Let $\rho =\frac {1}{2}\sum _{\alpha \in \Sigma ^+} m_\alpha \alpha $ .

The Cartan decomposition of G is given by

$$ \begin{align*} G=K(\exp \overline{\mathfrak{a}^+})K. \end{align*} $$

For $x \in G,$ we denote by $x^+ \in \overline {\mathfrak {a}^+}$ the corresponding component in the Cartan decomposition. We also have the Iwasawa decomposition given by $G=KAN$ and corresponding to the decomposition, we write for $g \in G$ , $g=k(g)\exp H(g) n(g)$ , where $k(g) \in K,\: n(g) \in N,$ and $H(g) \in \mathfrak {a}$ . The Haar measure on G relative to the Iwasawa decomposition $G=KAN$ is given by

(2.1) $$ \begin{align} \int_G f(g)\:dg \:=\: \int_K \int_{\mathfrak{a}} \int_N f\left(k(g)\exp\left(H(g)\right) n(g)\right)\:e^{2\rho(H(g))}\:dk\:dH\:dn, \end{align} $$

for all $f \in L^1(G)$ . For $g \in G$ , setting $\|g\|:=|g^+|$ , we have

(2.2) $$ \begin{align} \left|H(g)\right| \le \|g\|. \end{align} $$

Let M be the centralizer of A in K. For $\lambda \in \mathfrak {a}_{\mathbb {C}}$ , one has the spherical principal series representation $\pi _\lambda $ of G on $L^2(K/M)$ given by

$$ \begin{align*} \left(\pi_\lambda(x) V\right)(b):= e^{(i\lambda -\rho)H(x^{-1}b)}\:V(k(x^{-1}b)),\: \text{ for all } V \in L^2(K/M),\: b \in K \end{align*} $$

(see [Reference Gangolli and VaradarajanGV88] for more details). We consider an orthonormal basis $\{e_j\}_{j=0}^\infty $ of $L^2(K/M)$ with the unique K-fixed vector $e_0 \equiv 1$ . For $\lambda \in \mathfrak {a}$ , the Group Fourier transform of $f \in L^1(X)$ defined by

$$ \begin{align*} \widehat{f}(\pi_\lambda):=\int_G f(x)\:\pi_\lambda(x)\:dx \: \end{align*} $$

is a bounded linear operator on $L^2(K/M)$ and can be identified with the Helgason Fourier transform of f via

$$ \begin{align*} \widehat{f}(\pi_\lambda)(1)(b)=\int_G e^{(i\lambda - \rho)H(x^{-1}b)}\: f(x)\:dx = \tilde{f}(\lambda,b),\:\text{ for } b \in K/M. \end{align*} $$

The Helgason Fourier transform of f is injective on $L^1(G/K)$ , that is, if $\widehat {f}(\pi _\lambda )=0$ for all $\lambda \in \mathfrak {a}$ , then $f=0$ almost everywhere [Reference HelgasonHel08, Theorem 1.9, Chapter 3, p. 213].

For $\lambda \in \mathfrak {a}$ , the spherical function $\varphi _\lambda $ is given by

$$ \begin{align*} \varphi_\lambda(g)= \int_K e^{(-i\lambda + \rho)A(kg)}\: dk, \end{align*} $$

where $A(g)=-H(g^{-1})$ . This function is identified as the following matrix coefficient of the principal series representation $\pi _\lambda $ :

$$ \begin{align*} \varphi_\lambda(g) = \left\langle \pi_\lambda(g)e_0\:,\:e_0\right\rangle. \end{align*} $$

On $\mathfrak {a}$ , $\|\cdot \|$ denotes the norm inherited from $\mathfrak {p}$ . For $\lambda \in \mathfrak {a}$ , the spherical function $\varphi _\lambda $ is a smooth K-bi-invariant eigenfunction of the Laplace–Beltrami operator $\Delta $ :

$$ \begin{align*} \Delta \varphi_\lambda = -\left(\|\lambda\|^2 + \|\rho\|^2\right) \varphi_\lambda. \end{align*} $$

For $\lambda \in \mathfrak {a}$ , one has

(2.3) $$ \begin{align} |\varphi_\lambda(x)| \le 1,\:\text{ for all } x \in G. \end{align} $$

In fact, one has the following bounds which are decay estimates away from the identity [Reference Anker and JiAJ99, Proposition 2.2.12]:

(2.4) $$ \begin{align} |\varphi_\lambda(x)| \le \varphi_0(x) \lesssim \left\{\prod_{\alpha \in \Sigma^+_r} \left(1+\langle\alpha,x^+\rangle\right)\right\}e^{-\langle\rho,x^+\rangle},\:\text{ for all } x \in G. \end{align} $$

For $\lambda \in \mathfrak {a}$ and $f \in L^1(G//K)$ , the spherical Fourier transform of f,

$$ \begin{align*} \mathcal H f(\lambda)= \int_G f(x) \:\varphi_\lambda(x)\:dx, \end{align*} $$

is naturally identified with the matrix coefficient of the group Fourier transform given by $\left \langle\hspace{1pt} \widehat {f}\left (\pi _\lambda \right )e_0\:,\:e_0\right \rangle $ . $\mathcal H f$ is continuous on $\mathfrak {a}$ and also satisfies the Riemann–Lebesgue lemma [Reference HelgasonHel08, Theorem 1.8, Chapter 3, p. 209]:

(2.5) $$ \begin{align} \lim_{\|\lambda\| \to \infty} \mathcal H f(\lambda)=0. \end{align} $$

For $f \in C^\infty _c(G//K)$ , we have the Abel transform $\mathscr {A}f$ defined by

(2.6) $$ \begin{align} \mathscr{A}f(H)=e^{\rho(H)}\:\int_N f\left(\exp(H)n\right)\:dn. \end{align} $$

If $f \in L^1(G//K)$ , then $\mathscr {A}f \in L^1(\mathfrak {a})^W$ and moreover, by identifying $\mathfrak {a} \cong \mathfrak {a}^* \cong \mathbb R^d$ , where $d=rank(G)$ , we have for $\lambda \in \mathbb R^d$ ,

(2.7) $$ \begin{align} \mathscr F \left(\mathscr{A}f\right)(\lambda)= \mathcal H f(\lambda), \end{align} $$

where $\mathscr F$ is the classical Fourier transform on $\mathbb R^d$ .

Let $\mathcal {U}(\mathfrak {g})$ be the universal enveloping algebra of $\mathfrak {g}$ . The elements of $\mathcal {U}(\mathfrak {g})$ act on $C^\infty (G)$ as differential operators, on both sides. Following the notation of Harish-Chandra, we shall write $f\left (D;g;E\right )$ for the action of $(D,E) \in \mathcal {U}(\mathfrak {g}) \times \mathcal {U}(\mathfrak {g})$ on $f \in C^\infty (G)$ at $g \in G$ . More precisely,

$$ \begin{align*} f\left(D;g;E\right) &= \left(\frac{\partial}{\partial s_1} \cdots \frac{\partial}{\partial s_d}\frac{\partial}{\partial t_1}\cdots \frac{\partial}{\partial t_e}\right)\big{|}_{s_1=\cdots=s_d=t_1=\cdots=t_e=0} \\ &\quad \times f\left(\left(\exp s_1X_1\right)\cdots\left(\exp s_dX_d\right)g\left(\exp t_1Y_1\right)\cdots \left(\exp t_eY_e\right)\right), \end{align*} $$

StartLayout 1st Row 1st Column f left parenthesis upper D semicolon g semicolon upper E right parenthesis 2nd Column equals left parenthesis StartFraction partial differential Over partial differential s 1 EndFraction midline horizontal ellipsis StartFraction partial differential Over partial differential s Subscript d Baseline EndFraction StartFraction partial differential Over partial differential t 1 EndFraction midline horizontal ellipsis StartFraction partial differential Over partial differential t Subscript e Baseline EndFraction right parenthesis vertical bar Subscript s 1 equals midline horizontal ellipsis equals s Sub Subscript d Subscript equals t 1 equals midline horizontal ellipsis equals t Sub Subscript e Subscript equals 0 Baseline 2nd Row 1st Column Blank 2nd Column times f left parenthesis left parenthesis exp s 1 upper X 1 right parenthesis midline horizontal ellipsis left parenthesis exp s Subscript d Baseline upper X Subscript d Baseline right parenthesis g left parenthesis exp t 1 upper Y 1 right parenthesis midline horizontal ellipsis left parenthesis exp t Subscript e Baseline upper Y Subscript e Baseline right parenthesis right parenthesis comma EndLayout

if $D=X_1 \dots X_d$ and $E=Y_1 \dots Y_e$ , for $X_1, \ldots , X_d, Y_1, \ldots , Y_e \in \mathfrak {g}$ .

We now come to the definition of $L^1$ -Schwartz class functions. $\mathscr S^1(G)$ , called the $L^1$ -Schwartz class on G, is the collection of all $f \in C^\infty (G)$ such that for each $D,E \in \mathcal {U}(\mathfrak {g})$ and all nonnegative integers m,

$$ \begin{align*} \left|f\left(D;g;E\right)\right| \lesssim_m \varphi_0(g)^2 \left(1+\|g\|\right)^{-m},\: g \in G. \end{align*} $$

$\mathscr S^1(G//K)$ is the sub-collection of functions in $\mathscr S^1(G)$ that are also K-bi-invariant.

The $L^2$ -spectrum of $\Delta $ is given by $(-\infty ,-\|\rho \|^2]$ . We consider the shifted Laplace–Beltrami operator $\mathcal L:= \Delta + \|\rho \|^2$ , with $L^2$ -spectrum $(-\infty ,0]$ . Then for $\xi>0$ , the resolvent operator $\left (\xi I - \mathcal L\right )^{-1}$ is realized by convolution on the right with the K-bi-invariant tempered distribution $K_\xi $ on G given by [Reference AnkerAnk92, pp. 279–280]

$$ \begin{align*} K_\xi(x)= \int_0^\infty e^{-t\xi}\:h_t(x)\:dt, \end{align*} $$

where $h_t$ is the heat kernel on $\mathbb X$ .

We will also require the following result from complex analysis which is known as the Vivanti–Pringsheim theorem.

Lemma 2.1 [Reference RemmertRem91, p. 235]

Let the power series $g(z)=\sum _{\nu =0}^\infty b_{\nu }\:z^{\nu }$ have positive finite radius of convergence R and suppose that all but finitely many of its coefficients $b_{\nu }$ are real and nonnegative. Then $z_0:=R$ is a singular point of g.

3 Proofs of Theorem 1.1 and the Euclidean result

In this section, we prove Theorems 1.1 and 1.2. But first, we recall the definition of convolution.

On a second countable, locally compact, Hausdorff topological group G, equipped with a left Haar measure $\omega $ , the convolution of two functions $f_1,f_2 \in L^1(G)$ is defined as [Reference FollandFol15, p. 55]

$$ \begin{align*} f_1*f_2(x):=\int_G f_1(y) f_2(y^{-1}x)\: d\omega(y). \end{align*} $$

We now present the proof of Theorem 1.1.

Proof of Theorem 1.1

Let

$$ \begin{align*} N_1 = \min\{n \in [2,N] \mid a_n>0\}\:\:\text{ and }\:\:N_2 = \max\{n \in [2,N] \mid a_n>0\}. \end{align*} $$

We will inductively show that for each $j \in \mathbb N \cup \{0\}$ , there exists a sequence $\{m_{j,l}\}_{l=1}^\infty $ of nonnegative integers such that

(3.1) $$ \begin{align} \Psi_j = \sum_{l=1}^\infty m_{j,l} \left(*^l \psi\right), \end{align} $$ normal upper Psi Subscript j Baseline equals sigma summation Underscript l equals 1 Overscript infinity Endscripts m Subscript j comma l Baseline left parenthesis asterisk Superscript l Baseline psi right parenthesis comma

satisfying

(3.2) $$ \begin{align} m_{j,l}=0,\:\text{ for } l \ge N^j_2+1. \end{align} $$

To prove (3.2), we see that it is true for $j=0$ as

$$ \begin{align*} m_{0,l} = \begin{cases} 1, \text{ if } l=1, \\ 0, \text{ if } l\ge 2. \end{cases} \end{align*} $$ m Subscript 0 comma l Baseline equals StartLayout Enlarged left brace 1st Row 1 comma if l equals 1 comma 2nd Row 0 comma if l greater than or equals 2 period EndLayout

Let us assume (3.2) to be true for $j=k$ and then prove it for $j=k+1$ . By the iterative definition,

$$ \begin{align*} \Psi_{k+1} = \psi \:+\:\sum_{n=2}^N a_n \left(*^n \Psi_k\right). \end{align*} $$ normal upper Psi Subscript k plus 1 Baseline equals psi plus sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline left parenthesis asterisk Superscript n Baseline normal upper Psi Subscript k Baseline right parenthesis period

Now using the induction hypothesis, we can write

$$ \begin{align*} \Psi_{k+1} = \psi \:+\:\sum_{n=N_1}^{N_2} a_n \left[*^n \left\{\sum_{l=1}^{N^k_2}m_{k,l}\:\left(*^l \psi\right)\right\}\right]= \sum_{l=1}^{N^{k+1}_2} b_l \left(*^l \psi\right), \end{align*} $$

normal upper Psi Subscript k plus 1 Baseline equals psi plus sigma summation Underscript n equals upper N 1 Overscript upper N 2 Endscripts a Subscript n Baseline left bracket asterisk Superscript n Baseline StartSet sigma summation Underscript l equals 1 Overscript upper N 2 Superscript k Baseline Endscripts m Subscript k comma l Baseline left parenthesis asterisk Superscript l Baseline psi right parenthesis EndSet right bracket equals sigma summation Underscript l equals 1 Overscript upper N 2 Superscript k plus 1 Baseline Endscripts b Subscript l Baseline left parenthesis asterisk Superscript l Baseline psi right parenthesis comma

for some nonnegative integers $b_l$ for $l \in [1,N^{k+1}_2]$ . This gives (3.2).

We now claim that

(3.3) $$ \begin{align} m_{j+1,l} \ge m_{j,l}\:\text{ for all }l \in \mathbb N. \end{align} $$

To prove (3.3), we see that it is true for $l=1$ as $N_1 \ge 2$ and hence $m_{j,1}=1$ for all $j \in \mathbb N \cup \{0\}$ . Now assuming (3.3) to be true for $l \le k$ , we get (3.3) for $l=k+1$ by noting that

$$ \begin{align*} \Psi_{j+2} - \Psi_{j+1} = \sum_{n=N_1}^{N_2} a_n \sum_{p=0}^{n-1} \left(*^p \Psi_j\right) * \left(\Psi_{j+1}-\Psi_j\right) * \left(*^{n-p-1} \Psi_{j+1}\right) \end{align*} $$

and the induction hypothesis.

By (3.2) and (3.3), we get that $\{\Psi _j\}_{j=0}^\infty $ is an increasing sequence of nonnegative integrable functions on G. We next claim that

(3.4) $$ \begin{align} m_{j,l} = m_{j+1,l}\:\text{ for all }j \ge l. \end{align} $$

To prove (3.4), we note that

$$ \begin{align*} \Psi_1 - \Psi_0 = \sum_{n=N_1}^{N_2} a_n \left(*^n \psi\right), \end{align*} $$ normal upper Psi 1 minus normal upper Psi 0 equals sigma summation Underscript n equals upper N 1 Overscript upper N 2 Endscripts a Subscript n Baseline left parenthesis asterisk Superscript n Baseline psi right parenthesis comma

and for $j \ge 1$ ,

$$ \begin{align*} \Psi_{j+1} - \Psi_j &= \sum_{n=N_1}^{N_2} a_n \sum_{p=0}^{n-1} \left(*^p \Psi_{j-1}\right) * \left(\Psi_j-\Psi_{j-1}\right) * \left(*^{n-p-1} \Psi_j\right) \\ &= \left(\Psi_j-\Psi_{j-1}\right)*\sum_{n=N_1}^{N_2} a_n \sum_{p=0}^{n-1} \left(*^p \Psi_{j-1}\right) * \left(*^{n-p-1} \Psi_j\right) \\ &= \left(\Psi_j-\Psi_{j-1}\right)*\sum_{n=N_1}^{N_2} a_n \left\{n \left(*^{n-1} \psi \right) + \cdots\right\} \\ &= \left(\Psi_j-\Psi_{j-1}\right)* \left\{N_1a_{N_1}\left(*^{N_1-1} \psi \right)+\cdots\right\} \\ &= \left(\Psi_1-\Psi_0\right)* \left\{\left(N_1a_{N_1}\right)^{j-1}\left(*^{(N_1-1)(j-1)} \psi \right)+\cdots\right\} \\ &= \left\{a_{N_1}\left(*^{N_1} \psi \right)+\cdots\right\}* \left\{\left(N_1a_{N_1}\right)^{j-1}\left(*^{(N_1-1)(j-1)} \psi \right)+\cdots\right\} \\ &= N_1^{j-1}a^j_{N_1}\left(*^{(N_1j-j+1)} \psi \right)+\cdots. \end{align*} $$

StartLayout 1st Row 1st Column normal upper Psi Subscript j plus 1 Baseline minus normal upper Psi Subscript j 2nd Column equals sigma summation Underscript n equals upper N 1 Overscript upper N 2 Endscripts a Subscript n Baseline sigma summation Underscript p equals 0 Overscript n minus 1 Endscripts left parenthesis asterisk Superscript p Baseline normal upper Psi Subscript j minus 1 Baseline right parenthesis asterisk left parenthesis normal upper Psi Subscript j Baseline minus normal upper Psi Subscript j minus 1 Baseline right parenthesis asterisk left parenthesis asterisk Superscript n minus p minus 1 Baseline normal upper Psi Subscript j Baseline right parenthesis 2nd Row 1st Column Blank 2nd Column equals left parenthesis normal upper Psi Subscript j Baseline minus normal upper Psi Subscript j minus 1 Baseline right parenthesis asterisk sigma summation Underscript n equals upper N 1 Overscript upper N 2 Endscripts a Subscript n Baseline sigma summation Underscript p equals 0 Overscript n minus 1 Endscripts left parenthesis asterisk Superscript p Baseline normal upper Psi Subscript j minus 1 Baseline right parenthesis asterisk left parenthesis asterisk Superscript n minus p minus 1 Baseline normal upper Psi Subscript j Baseline right parenthesis 3rd Row 1st Column Blank 2nd Column equals left parenthesis normal upper Psi Subscript j Baseline minus normal upper Psi Subscript j minus 1 Baseline right parenthesis asterisk sigma summation Underscript n equals upper N 1 Overscript upper N 2 Endscripts a Subscript n Baseline left brace n left parenthesis asterisk Superscript n minus 1 Baseline psi right parenthesis plus midline horizontal ellipsis right brace 4th Row 1st Column Blank 2nd Column equals left parenthesis normal upper Psi Subscript j Baseline minus normal upper Psi Subscript j minus 1 Baseline right parenthesis asterisk left brace upper N 1 a Subscript upper N 1 Baseline left parenthesis asterisk Superscript upper N 1 minus 1 Baseline psi right parenthesis plus midline horizontal ellipsis right brace 5th Row 1st Column Blank 2nd Column equals left parenthesis normal upper Psi 1 minus normal upper Psi 0 right parenthesis asterisk left brace left parenthesis upper N 1 a Subscript upper N 1 Baseline right parenthesis Superscript j minus 1 Baseline left parenthesis asterisk Superscript left parenthesis upper N 1 minus 1 right parenthesis left parenthesis j minus 1 right parenthesis Baseline psi right parenthesis plus midline horizontal ellipsis right brace 6th Row 1st Column Blank 2nd Column equals left brace a Subscript upper N 1 Baseline left parenthesis asterisk Superscript upper N 1 Baseline psi right parenthesis plus midline horizontal ellipsis right brace asterisk left brace left parenthesis upper N 1 a Subscript upper N 1 Baseline right parenthesis Superscript j minus 1 Baseline left parenthesis asterisk Superscript left parenthesis upper N 1 minus 1 right parenthesis left parenthesis j minus 1 right parenthesis Baseline psi right parenthesis plus midline horizontal ellipsis right brace 7th Row 1st Column Blank 2nd Column equals upper N 1 Superscript j minus 1 Baseline a Subscript upper N 1 Superscript j Baseline left parenthesis asterisk Superscript left parenthesis upper N 1 j minus j plus 1 right parenthesis Baseline psi right parenthesis plus midline horizontal ellipsis period EndLayout

Now as $N_1 \ge 2$ , (3.4) follows.

From (3.4), it follows that $\Psi _j \uparrow \Psi $ , where

(3.5) $$ \begin{align} \Psi = \sum_{l=1}^\infty m_{l,l} \left(*^l \psi\right). \end{align} $$ normal upper Psi equals sigma summation Underscript l equals 1 Overscript infinity Endscripts m Subscript l comma l Baseline left parenthesis asterisk Superscript l Baseline psi right parenthesis period

Finally, we will show that

(3.6) $$ \begin{align} \|\Psi_j\|_{L^1(G)} = \sum_{l=1}^{N^j_2} m_{j,l}\:\|\psi\|^l_{L^1(G)}\le \sum_{l=1}^{N^j_2} m_{j,l}\:\mathcal Q(t_{\mathcal Q})^l \le t_{\mathcal Q}. \end{align} $$

Assuming (3.6), we see that the series in (3.5) converges absolutely. Then the desired properties of $\Psi $ follow. So we are left to prove (3.6). To show (3.6), we note that for $j=0$ , the assertion is true as by (1.6), it follows that

$$ \begin{align*} \|\Psi_0\|_{L^1(G)} = \|\psi\|_{L^1(G)} \le \mathcal Q(t_{\mathcal Q}) < t_{\mathcal Q}. \end{align*} $$

For general $j \ge 1$ , by construction and the hypothesis, the equality and the first inequality in (3.6) follow. So we assume

(3.7) $$ \begin{align} \sum_{l=1}^{N^j_2} m_{j,l}\:\mathcal Q(t_{\mathcal Q})^l \le t_{\mathcal Q}\: \end{align} $$ sigma summation Underscript l equals 1 Overscript upper N 2 Superscript j Baseline Endscripts m Subscript j comma l Baseline script upper Q left parenthesis t Subscript script upper Q Baseline right parenthesis Superscript l Baseline less than or equals t Subscript script upper Q

to be true for $j=k$ and show it to be true for $j=k+1$ . Now consider a nonnegative $\psi \in L^1(G)$ with $\int _G\psi = \mathcal Q\left (t_{\mathcal Q}\right )$ . Then defining $\{\Psi _j\}_{j=0}^\infty $ iteratively and then integrating out the relation,

$$ \begin{align*} \Psi_{k+1} = \psi \:+\:\sum_{n=2}^N a_n \left(*^n \Psi_k\right), \end{align*} $$ normal upper Psi Subscript k plus 1 Baseline equals psi plus sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline left parenthesis asterisk Superscript n Baseline normal upper Psi Subscript k Baseline right parenthesis comma

with respect to the fixed left Haar measure, we get that

(3.8) $$ \begin{align} \int_G \Psi_{k+1} = \int_G \psi +\sum_{n=2}^N a_n \left(\int_G \Psi_k\right)^n. \end{align} $$ integral Underscript upper G Endscripts normal upper Psi Subscript k plus 1 Baseline equals integral Underscript upper G Endscripts psi plus sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline left parenthesis integral Underscript upper G Endscripts normal upper Psi Subscript k Baseline right parenthesis Superscript n Baseline period

Now using (3.1) and (3.2), we get

(3.9) $$ \begin{align} \int_G \Psi_{k+1} = \sum_{l=1}^{N^{k+1}_2} m_{k+1,l}\: \left(\int_G \psi\right)^l = \sum_{l=1}^{N^{k+1}_2} m_{k+1,l}\: \mathcal Q(t_{\mathcal Q})^l, \end{align} $$

and also

(3.10) $$ \begin{align} \sum_{n=2}^N a_n \left(\int_G \Psi_k\right)^n &= \sum_{n=2}^N a_n \left[\int_G \left\{\sum_{l=1}^{N^k_2} m_{k,l} \left(*^l \psi\right)\right\}\right]^n \nonumber\\ &= \sum_{n=2}^N a_n \left[\sum_{l=1}^{N^k_2} m_{k,l} \left(\int_G \psi\right)^l\right]^n \nonumber\\ &= \sum_{n=2}^N a_n \left(\sum_{l=1}^{N^k_2} m_{k,l} \mathcal Q(t_{\mathcal Q})^l\right)^n. \end{align} $$

StartLayout 1st Row 1st Column sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline left parenthesis integral Underscript upper G Endscripts normal upper Psi Subscript k Baseline right parenthesis Superscript n 2nd Column equals sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline left bracket integral Underscript upper G Endscripts StartSet sigma summation Underscript l equals 1 Overscript upper N 2 Superscript k Baseline Endscripts m Subscript k comma l Baseline left parenthesis asterisk Superscript l Baseline psi right parenthesis EndSet right bracket Superscript n Baseline 2nd Row 1st Column Blank 2nd Column equals sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline left bracket sigma summation Underscript l equals 1 Overscript upper N 2 Superscript k Baseline Endscripts m Subscript k comma l Baseline left parenthesis integral Underscript upper G Endscripts psi right parenthesis Superscript l Baseline right bracket Superscript n Baseline 3rd Row 1st Column Blank 2nd Column equals sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline left parenthesis sigma summation Underscript l equals 1 Overscript upper N 2 Superscript k Baseline Endscripts m Subscript k comma l Baseline script upper Q left parenthesis t Subscript script upper Q Baseline right parenthesis Superscript l Baseline right parenthesis Superscript n Baseline period EndLayout

Then plugging (3.9) and (3.10) in (3.8) and using (3.7) for $j=k$ along with the fact that the polynomial $\mathcal P(t):=\sum _{n=2}^N a_n\:t^n$ is strictly increasing on $(0,\infty )$ , it follows that

$$ \begin{align*} \sum_{l=1}^{N^{k+1}_2} m_{k+1,l}\: \mathcal Q(t_{\mathcal Q})^l = \mathcal Q(t_{\mathcal Q}) + \sum_{n=2}^N a_n \left(\sum_{l=1}^{N^k_2} m_{k,l} \mathcal Q(t_{\mathcal Q})^l\right)^n \le \mathcal Q(t_{\mathcal Q}) + \sum_{n=2}^N a_n\:t^n_{\mathcal Q} = t_{\mathcal Q}. \end{align*} $$

StartLayout 1st Row sigma summation Underscript l equals 1 Overscript upper N 2 Superscript k plus 1 Baseline Endscripts m Subscript k plus 1 comma l Baseline script upper Q left parenthesis t Subscript script upper Q Baseline right parenthesis Superscript l Baseline equals script upper Q left parenthesis t Subscript script upper Q Baseline right parenthesis plus sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline left parenthesis sigma summation Underscript l equals 1 Overscript upper N 2 Superscript k Baseline Endscripts m Subscript k comma l Baseline script upper Q left parenthesis t Subscript script upper Q Baseline right parenthesis Superscript l Baseline right parenthesis Superscript n Baseline less than or equals script upper Q left parenthesis t Subscript script upper Q Baseline right parenthesis plus sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline t Subscript script upper Q Superscript n Baseline equals t Subscript script upper Q Baseline period EndLayout

This completes the proof of (3.6) and hence also of Theorem 1.1.

We now present the proof of Theorem 1.2.

Proof of Theorem 1.2

Let f be as in the hypothesis of Theorem 1.2. We now define

(3.11) $$ \begin{align} \psi := f - \sum_{n=2}^N a_n \left(*^n f\right). \end{align} $$ psi colon equals f minus sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline left parenthesis asterisk Superscript n Baseline f right parenthesis period

Then $\psi \in L^1(\mathbb R^d)$ is nonnegative. Moreover, keeping in mind the hypothesis that $\int _{\mathbb R^d} f \ge 0$ , we integrate the relation (3.11) and then use the fact that $\mathcal Q$ attains a unique positive maximum on $(0,\infty )$ at $t_{\mathcal Q}$ to obtain

$$ \begin{align*} 0 \le \int_{\mathbb R^d} \psi = \int_{\mathbb R^d} f - \sum_{n=2}^N a_n \left(\int_{\mathbb R^d} f\right)^n = \mathcal Q\left(\int_{\mathbb R^d} f\right) \le \mathcal Q(t_{\mathcal Q}). \end{align*} $$

Hence, we also get

$$ \begin{align*} \|\psi\|_{L^1(\mathbb R^d)} \le \mathcal Q(t_{\mathcal Q}). \end{align*} $$

Then applying Theorem 1.1, we get a $\Psi \in L^1(\mathbb R^d)$ such that

(3.12) $$ \begin{align} \Psi \ge 0\:,\: \|\Psi\|_{L^1(\mathbb R^d)} \le t_{\mathcal Q}\: \text{ and } \psi = \Psi - \sum_{n=2}^N a_n \left(*^n \Psi\right). \end{align} $$

Now if we can show that the Fourier transforms $\mathscr F f$ and $\mathscr F \Psi $ agree, that is,

(3.13) $$ \begin{align} \mathscr F f(\xi)= \mathscr F \Psi(\xi)\:,\: \text { for all } \xi \in \mathbb R^d, \end{align} $$

then by injectivity of the Fourier transform, it will follow that $f=\Psi $ as elements of $L^1(\mathbb R^d)$ and f will inherit the desired properties from $\Psi $ stated in (3.12), which will then complete the proof of Theorem 1.2. Thus, it is enough to prove (3.13).

As $\Psi $ is nonnegative, we claim that

(3.14) $$ \begin{align} \left|\mathscr F \Psi(\xi)\right| < t_{\mathcal Q},\:\text{ for all } \xi \in \mathbb R^d \setminus \{0\}. \end{align} $$

If $\Psi $ is a trivial element in $L^1(\mathbb R^d)$ , that is, $\Psi $ is zero almost everywhere, then trivially

$$ \begin{align*} \left|\mathscr F \Psi(\xi)\right|=0< t_{\mathcal Q},\:\text{ for all } \xi \in \mathbb R^d. \end{align*} $$

To see (3.14) for $\Psi $ nontrivial, by nonnegativity of $\Psi $ , we have for all $\xi \in \mathbb R^d$

(3.15) $$ \begin{align} \left|\mathscr F \Psi(\xi)\right| = \left|\int_{\mathbb R^d} \Psi(x)\:e^{-2\pi ix\cdot\xi}\:dx\right| \le \int_{\mathbb R^d} \Psi(x)\:dx. \end{align} $$

We now find out for which $\xi $ , one can actually expect to get equality in (3.15). The equality above and the fact that $\Psi>0$ on a set of positive Lebesgue measure would yield a constant $\alpha \in {\mathbb {C}}$ such that

$$ \begin{align*} \alpha\:e^{-2\pi ix\cdot\xi}=1\: \end{align*} $$

holds on a set of positive Lebesgue measure and hence, by real-analyticity, actually holds on $\mathbb R^d$ . Then for $1\le j \le d$ , considering jth partial derivatives $\frac {\partial }{\partial x_j}$ , we get

$$ \begin{align*} -2\pi i \xi_j\: e^{-2\pi ix\cdot\xi}=0, \end{align*} $$

which implies $\xi _j=0$ for all $1\le j \le d$ . Thus, (3.15) is strict inequality unless $\xi =0$ . Then by (3.12),

$$ \begin{align*} \left|\mathscr F \Psi(\xi)\right| < \int_{\mathbb R^d} \Psi \le t_{\mathcal Q},\:\text{ for all } \xi \in \mathbb R^d \setminus \{0\}. \end{align*} $$

This proves the claim (3.14).

We now define

$$ \begin{align*} \Omega:= \left\{\xi \in \mathbb R^d \setminus \{0\} \mid \mathscr F f(\xi)= \mathscr F \Psi(\xi)\right\}\subset \mathbb R^d \setminus \{0\}, \end{align*} $$

and by adopting a topological argument, we show that $\Omega = \mathbb R^d \setminus \{0\}$ . Since both $\mathscr F f$ and $\mathscr F \Psi $ are continuous, it follows that $\Omega $ is closed in $\mathbb R^d \setminus \{0\}$ . To show that $\Omega $ is also open in $\mathbb R^d \setminus \{0\}$ , we let $\xi _0 \in \Omega $ . Then by (3.14), there exists $\varepsilon>0$ such that

(3.16) $$ \begin{align} \left|\mathscr F \Psi(\xi_0)\right| < (1-\varepsilon)t_{\mathcal Q}. \end{align} $$

By continuity of $\mathscr F f$ , there exists $\delta>0$ such that

(3.17) $$ \begin{align} \left|\left|\mathscr F f(\xi_0)\right|-\left|\mathscr F f(\xi)\right|\right| < \varepsilon\: t_{\mathcal Q}\:,\:\text{ for all } \xi \in B(\xi_0,\delta), \end{align} $$

where $B(\xi _0,\delta ) \subset \mathbb R^d \setminus \{0\}$ is the Euclidean ball with center $\xi _0$ and radius $\delta $ . Now as $\xi _0 \in \Omega $ , by (3.16), we have

$$ \begin{align*} \left|\mathscr F f(\xi_0)\right|=\left|\mathscr F \Psi(\xi_0)\right| < (1-\varepsilon)t_{\mathcal Q}, \end{align*} $$

which combined with (3.17) yields

$$ \begin{align*} \left|\mathscr F f(\xi)\right| < t_{\mathcal Q},\:\text{ for all } \xi \in B(\xi_0,\delta). \end{align*} $$

So, in particular, both

(3.18) $$ \begin{align} \left|\mathscr F \Psi(\xi)\right|<t_{\mathcal Q} \:\text{ and }\:\left|\mathscr F f(\xi)\right| <t_{\mathcal Q},\:\text{ for all } \xi \in B(\xi_0,\delta). \end{align} $$

Now as by (3.11) and (3.12),

$$ \begin{align*} f - \sum_{n=2}^N a_n \left(*^n f\right) = \psi = \Psi - \sum_{n=2}^N a_n \left(*^n \Psi\right), \end{align*} $$

by taking the Fourier transform, we have

(3.19) $$ \begin{align} \mathscr F f - \sum_{n=2}^N a_n \left(\mathscr F f\right)^n = \mathscr F \Psi - \sum_{n=2}^N a_n \left(\mathscr F \Psi\right)^n. \end{align} $$

Then in view of (3.18), we consider the complex polynomial $\mathcal Q : {\mathbb {C}} \to {\mathbb {C}}$ and the disk

$$ \begin{align*} \mathcal{D} := \left\{z \in {\mathbb{C}} \mid |z|< t_{\mathcal Q}\right\}. \end{align*} $$

We claim that $\mathcal Q$ is injective on $\mathcal {D}$ . To prove the claim, we write

$$ \begin{align*} \mathcal Q(z) = z - \sum_{n=2}^N a_n z^n = z - \mathcal P(z). \end{align*} $$ script upper Q left parenthesis z right parenthesis equals z minus sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline z Superscript n Baseline equals z minus script upper P left parenthesis z right parenthesis period

Then

$$ \begin{align*} \mathcal Q'(z) = 1 - \sum_{n=2}^N na_n z^{n-1} = 1 - \mathcal P'(z), \end{align*} $$ script upper Q prime left parenthesis z right parenthesis equals 1 minus sigma summation Underscript n equals 2 Overscript upper N Endscripts n a Subscript n Baseline z Superscript n minus 1 Baseline equals 1 minus script upper P prime left parenthesis z right parenthesis comma

and hence $\mathcal Q'(0)=1$ and $\mathcal P'(0)=0$ . Now as $\mathcal Q'$ is a polynomial in one complex variable, by the fundamental theorem of algebra, it has finitely many zeros. Let $z_0$ be a zero of $\mathcal Q'$ of smallest modulus. We now consider the disk

$$ \begin{align*} \mathcal{D}' := \left\{ z \in {\mathbb{C}} \mid |z| < |z_0| \right\}. \end{align*} $$

Now as $z_0$ is a zero of $\mathcal Q'$ of smallest modulus, $\mathcal P'(z) \ne 1$ for all $z \in \mathcal {D}'$ . We focus on the line segment $(0,|z_0|) \subset \mathcal {D}' \cap (0,\infty )$ . As the coefficients $a_n$ are nonnegative with at least one strictly positive, it follows that $\mathcal P'$ is strictly increasing on $(0,|z_0|)$ . Combining it with the (already noted) facts that $\mathcal P'(0)=0$ and $\mathcal P'(z) \ne 1$ for all $z \in \mathcal {D}'$ , it follows that

(3.20) $$ \begin{align} \mathcal P'(z)< 1 \:,\text{ for } z \in (0,|z_0|). \end{align} $$

Then by nonnegativity of the coefficients $a_n$ and (3.20), it follows that for all $z \in \mathcal {D}'$ ,

$$ \begin{align*} \left|\mathcal P'(z)\right| \le \sum_{n=2}^N na_n |z|^{n-1} = \mathcal P'(|z|) < 1. \end{align*} $$

Then by the geometric series expansion, we can write $1/\mathcal Q'$ in the form of a convergent power series in $\mathcal {D}'$ ,

$$ \begin{align*} \frac{1}{\mathcal Q'(z)}= \frac{1}{1 - \mathcal P'(z)}= 1 + \sum_{n=1}^\infty \left(\mathcal P'(z)\right)^n, \end{align*} $$

with real and nonnegative coefficients and radius of convergence $|z_0|$ . Then by the Vivanti–Pringsheim theorem (Lemma 2.1), $|z_0|$ is a singularity of $1/\mathcal Q'$ . Hence, $|z_0| \in (0,\infty )$ is a zero of $\mathcal Q'$ . But since $t_{\mathcal Q}$ is the unique zero of $\mathcal Q'$ in $(0,\infty )$ , it follows that $|z_0|=t_{\mathcal Q}$ and $\mathcal {D}=\mathcal {D}'$ . Consequently, we get that

(3.21) $$ \begin{align} \left|\mathcal P'(z)\right| < 1,\text{ for all } z \in \mathcal{D}. \end{align} $$

Now for two distinct $z_1,z_2 \in \mathcal {D}$ , we have

$$ \begin{align*} \mathcal Q(z_1)-\mathcal Q(z_2) = (z_1-z_2) -\int_{z_2}^{z_1} \mathcal P'(\xi)\:d\xi, \end{align*} $$

where the above line integral is along the line segment joining $z_2$ to $z_1$ . Then by triangle inequality and (3.21), it follows that

$$ \begin{align*} \left|\mathcal Q(z_1)-\mathcal Q(z_2) \right| \ge \left|z_1-z_2\right| -\left|\int_{z_2}^{z_1} \mathcal P'(\xi)\:d\xi \right|> \left|z_1-z_2\right| - \left|z_1-z_2\right| =0. \end{align*} $$

This proves the claim that $\mathcal Q$ is injective on $\mathcal {D}$ .

Now (3.18) can be rewritten as

$$ \begin{align*} \mathscr F \Psi(\xi)\:,\:\mathscr F f(\xi) \in \mathcal{D},\:\text{ for all } \xi \in B(\xi_0,\delta), \end{align*} $$

which along with (3.19) and the injectivity of $\mathcal Q$ on $\mathcal {D}$ yield

$$ \begin{align*} \mathscr F \Psi(\xi)=\mathscr F f(\xi),\:\text{ for all } \xi \in B(\xi_0,\delta), \end{align*} $$

that is, $B(\xi _0,\delta ) \subset \Omega $ and hence $\Omega $ is also open in $\mathbb R^d \setminus \{0\}$ .

We finally show that $\Omega $ is non-empty. By the Riemann–Lebesgue lemma, there exists $\xi _0 \in \mathbb R^d \setminus \{0\}$ such that

$$ \begin{align*} \left|\mathscr F f(\xi_0)\right| < t_{\mathcal Q}. \end{align*} $$

Thus, by (3.14), we get that both $\mathscr F f(\xi _0), \mathscr F \Psi (\xi _0) \in \mathcal {D}$ and then by (3.19) and the injectivity of $\mathcal Q$ on $\mathcal {D}$ , we obtain $\mathscr F f(\xi _0)= \mathscr F \Psi (\xi _0)$ , that is, $\xi _0 \in \Omega $ and hence $\Omega $ is non-empty. Moreover, for the special when $d=1$ , implementing the above argument involving the Riemann–Lebesgue lemma on both the connected components of $\mathbb R \setminus \{0\}$ , we can also see that $\Omega $ has non-empty intersections with both the connected components of $\mathbb R \setminus \{0\}$ .

For $d \ge 2$ , since $\Omega $ is a non-empty subset of $\mathbb R^d \setminus \{0\}$ which is both closed as well as open, we have $\Omega =\mathbb R^d \setminus \{0\}$ and thus (3.13) is true for $\mathbb R^d \setminus \{0\}$ . For $d=1$ , $\Omega $ is both closed as well as open in $\mathbb R \setminus \{0\}$ and, moreover, has non-empty intersections with both the connected components of $\mathbb R \setminus \{0\}$ and thus (3.13) is true for $\mathbb R \setminus \{0\}$ . The equality at $0$ then follows by continuity. This completes the proof of Theorem 1.2.

Remark 3.1

(1) The nonnegativity of the integral condition in Theorem 1.2 is necessary. For the counter-example, we refer to [Reference Nakamura and SawanoNS25, point (2) of Remark $1.3$ ].
(2) The analogue of Theorem 1.2 fails for the unit circle $\mathbb T$ which is identified with $[-\pi ,\pi ]$ equipped with the normalized Lebesgue measure $\frac {dx}{2\pi }$ . On $\mathbb T$ , we consider the trigonometric polynomial
$$ \begin{align*} f(x):= 1+2\cos(x). \end{align*} $$
Then $\int _{\mathbb T} f = 1$ and the Fourier transform of f is given by
$$ \begin{align*} \widehat{f}(n)= \begin{cases} 1\:, \text{ if } n \in \{-1,0,1\}, \\ 0\:, \text{ if } n \in \mathbb Z \setminus \{-1,0,1\}, \end{cases} \end{align*} $$
which yields $\widehat {f}(n)=\widehat {f}(n)^2$ for all $n \in \mathbb Z$ and thus $f=f*f$ on $\mathbb T$ . But f assumes negative values on $[-\pi ,-\frac {2\pi }{3}) \cup (\frac {2\pi }{3},\pi ]$ and a simple computation yields
$$ \begin{align*} \|f\|_{L^1(\mathbb T)}= \left(\frac{1}{3}+\frac{2\sqrt{3}}{\pi}\right)>1. \end{align*} $$

4 Results on symmetric spaces

In this section, we prove Theorems 1.4, 1.5, and 1.7. We first present the proof of Theorem 1.4.

Proof of Theorem 1.4

Let f be as in the hypothesis of Theorem 1.4. We now define

(4.1) $$ \begin{align} \psi := f - \sum_{n=2}^N a_n \left(*^n f\right). \end{align} $$ psi colon equals f minus sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline left parenthesis asterisk Superscript n Baseline f right parenthesis period

Then $\psi \in L^1(G/K)$ is nonnegative. Moreover, keeping in mind the hypothesis that $\int _G f \ge 0$ , integrating the relation (4.1) and then using the properties of $t_{\mathcal Q}$ , we again get

$$ \begin{align*} \|\psi\|_{L^1(G)} \le \mathcal Q(t_{\mathcal Q}). \end{align*} $$

Then applying Theorem 1.1, we get a $\Psi \in L^1(G/K)$ such that

(4.2) $$ \begin{align} \Psi \ge 0\:,\: \|\Psi\|_{L^1(G)} \le t_{\mathcal Q}\: \text{ and } \psi = \Psi - \sum_{n=2}^N a_n \left(*^n \Psi\right). \end{align} $$

Now by injectivity of the Helgason Fourier transform, it suffices to show that

(4.3) $$ \begin{align} \widehat f(\pi_\lambda)= \widehat \Psi(\pi_\lambda),\: \text { for all } \lambda \in \mathfrak{a}, \end{align} $$

as then it will follow that $f=\Psi $ as elements of $L^1(G/K)$ and f will inherit the desired properties from $\Psi $ stated in (4.2), which will then complete the proof of Theorem 1.4. Thus, it is enough to prove (4.3).

By (4.1) and (4.2), we have

$$ \begin{align*} f - \sum_{n=2}^N a_n \left(*^n f\right)= \psi = \Psi - \sum_{n=2}^N a_n \left(*^n \Psi\right), \end{align*} $$

which upon taking the group Fourier transform becomes

(4.4) $$ \begin{align} \widehat f(\pi_\lambda) - \sum_{n=2}^N a_n \widehat f(\pi_\lambda)^n= \widehat \Psi(\pi_\lambda) - \sum_{n=2}^N a_n \widehat \Psi(\pi_\lambda)^n. \end{align} $$

The equality above is of bounded operators on the Hilbert space $L^2(K/M)$ . We consider an orthonormal basis $\{e_j\}_{j=0}^\infty $ of $L^2(K/M)$ with the unique K-fixed vector $e_0 \equiv 1$ . Now for $(p,q) \in \left (\mathbb N \cup \{0\}\right )^2$ , the equality (4.4) yields the corresponding equality in matrix coefficients

(4.5) $$ \begin{align} \widehat f(\pi_\lambda)_{p,q} - \sum_{n=2}^N a_n \left(\widehat f(\pi_\lambda)^n\right)_{p,q}= \widehat \Psi(\pi_\lambda)_{p,q} - \sum_{n=2}^N a_n \left(\widehat \Psi(\pi_\lambda)^n\right)_{p,q}, \end{align} $$

where

$$ \begin{align*} \widehat f(\pi_\lambda)_{p,q} = \left\langle \widehat f(\pi_\lambda)e_q\:,\:e_p \right\rangle, \end{align*} $$

and so on. We first note that as f is right K-invariant,

(4.6) $$ \begin{align} \widehat f(\pi_\lambda)_{p,q} = 0\:,\: \text{ if } (p,q) \in \left(\mathbb N \cup \{0\}\right) \times \mathbb N, \end{align} $$

and similarly for $\Psi $ . We now claim that for all $n \ge 2$ ,

(4.7) $$ \begin{align} \left(\hspace{1.5pt}\widehat f(\pi_\lambda)^n\right)_{p,q} = \widehat f(\pi_\lambda)_{p,q}\:\widehat f(\pi_\lambda)^{n-1}_{0,0}. \end{align} $$

We prove the above claim by induction. For $n=2$ ,

$$ \begin{align*} \left(\hspace{1.5pt}\widehat f(\pi_\lambda)^2\right)_{p,q} = \left\langle\hspace{1.5pt} \widehat f(\pi_\lambda)\left(\hspace{1.5pt}\widehat f(\pi_\lambda)e_q\right)\:,\:e_p \right\rangle &= \left\langle \hspace{1.5pt} \widehat f(\pi_\lambda)\left(\sum_{r=0}^\infty \left\langle\hspace{1.5pt}\widehat f(\pi_\lambda)e_q\:,\:e_r\right\rangle e_r \right)\:,\:e_p \right\rangle \\ &= \sum_{r=0}^\infty \left\langle\hspace{1.5pt}\widehat f(\pi_\lambda)e_q\:,\:e_r\right\rangle \left\langle\hspace{1.5pt}\widehat f(\pi_\lambda)e_r\:,\:e_p\right\rangle \\ &= \sum_{r=0}^\infty \widehat f(\pi_\lambda)_{r,q}\:\widehat f(\pi_\lambda)_{p,r}. \end{align*} $$

StartLayout 1st Row 1st Column left parenthesis ModifyingAbove f With caret left parenthesis pi Subscript lamda Baseline right parenthesis squared right parenthesis Subscript p comma q Baseline equals left angle bracket ModifyingAbove f With caret left parenthesis pi Subscript lamda Baseline right parenthesis left parenthesis ModifyingAbove f With caret left parenthesis pi Subscript lamda Baseline right parenthesis e Subscript q Baseline right parenthesis comma e Subscript p Baseline right angle bracket 2nd Column equals left angle bracket ModifyingAbove f With caret left parenthesis pi Subscript lamda Baseline right parenthesis left parenthesis sigma summation Underscript r equals 0 Overscript infinity Endscripts left angle bracket ModifyingAbove f With caret left parenthesis pi Subscript lamda Baseline right parenthesis e Subscript q Baseline comma e Subscript r Baseline right angle bracket e Subscript r Baseline right parenthesis comma e Subscript p Baseline right angle bracket 2nd Row 1st Column Blank 2nd Column equals sigma summation Underscript r equals 0 Overscript infinity Endscripts left angle bracket ModifyingAbove f With caret left parenthesis pi Subscript lamda Baseline right parenthesis e Subscript q Baseline comma e Subscript r Baseline right angle bracket left angle bracket ModifyingAbove f With caret left parenthesis pi Subscript lamda Baseline right parenthesis e Subscript r Baseline comma e Subscript p Baseline right angle bracket 3rd Row 1st Column Blank 2nd Column equals sigma summation Underscript r equals 0 Overscript infinity Endscripts ModifyingAbove f With caret left parenthesis pi Subscript lamda Baseline right parenthesis Subscript r comma q Baseline ModifyingAbove f With caret left parenthesis pi Subscript lamda Baseline right parenthesis Subscript p comma r Baseline period EndLayout

Thus, if $q \in \mathbb N$ , by (4.6) and the above,

$$ \begin{align*} \left(\hspace{1.5pt}\widehat f(\pi_\lambda)^2\right)_{p,q} = 0 = \widehat f(\pi_\lambda)_{p,q}\:\widehat f(\pi_\lambda)_{0,0}. \end{align*} $$

For $q=0$ , again by (4.6) and the above,

$$ \begin{align*} \left(\hspace{1.5pt}\widehat f(\pi_\lambda)^2\right)_{p,0} = \widehat f(\pi_\lambda)_{0,0}\:\widehat f(\pi_\lambda)_{p,0}, \end{align*} $$

and hence (4.7) is established for $n=2$ .

Now assuming (4.7) for $n=j$ , we see that

$$ \begin{align*} \left(\hspace{1.5pt}\widehat f(\pi_\lambda)^{j+1}\right)_{p,q} = \left\langle \widehat f(\pi_\lambda)\left(\hspace{1.5pt}\widehat f(\pi_\lambda)^j e_q\right)\:,\:e_p \right\rangle &= \left\langle\hspace{1.5pt} \widehat f(\pi_\lambda)\left(\sum_{r=0}^\infty \left\langle\hspace{1.5pt}\widehat f(\pi_\lambda)^j e_q\:,\:e_r\right\rangle e_r \right)\:,\:e_p \right\rangle \\ &= \sum_{r=0}^\infty \left\langle\hspace{1.5pt}\widehat f(\pi_\lambda)^j e_q\:,\:e_r\right\rangle \left\langle\hspace{1.5pt}\widehat f(\pi_\lambda)e_r\:,\:e_p\right\rangle \\ &= \sum_{r=0}^\infty \left(\hspace{1.5pt}\widehat f(\pi_\lambda)^j\right)_{r,q}\:\widehat f(\pi_\lambda)_{p,r}\\ &= \sum_{r=0}^\infty \widehat f(\pi_\lambda)_{r,q}\:\widehat f(\pi_\lambda)^{j-1}_{0,0}\:\widehat f(\pi_\lambda)_{p,r}. \end{align*} $$

Now if $q \in \mathbb N$ , by (4.6),

$$ \begin{align*} \left(\hspace{1.5pt}\widehat f(\pi_\lambda)^{j+1}\right)_{p,q} = 0 = \widehat f(\pi_\lambda)_{p,q}\:\widehat f(\pi_\lambda)^j_{0,0}, \end{align*} $$

and for $q=0$ ,

$$ \begin{align*} \left(\hspace{1.5pt}\widehat f(\pi_\lambda)^{j+1}\right)_{p,0} = \widehat f(\pi_\lambda)^j_{0,0}\:\widehat f(\pi_\lambda)_{p,0}. \end{align*} $$

This proves the claim (4.7). Similar result is true for $\Psi $ as well.

Now noting (4.6) and plugging (4.7) in (4.5), we obtain for $p \in \mathbb N \cup \{0\}$

(4.8) $$ \begin{align} \widehat f(\pi_\lambda)_{p,0}\left(1 - \sum_{n=2}^N a_n \widehat f(\pi_\lambda)^{n-1}_{0,0}\right)= \widehat \Psi(\pi_\lambda)_{p,0}\left(1 - \sum_{n=2}^N a_n \widehat \Psi(\pi_\lambda)^{n-1}_{0,0}\right). \end{align} $$

For $p=0$ , (4.8) turns out to be

(4.9) $$ \begin{align} \widehat f(\pi_\lambda)_{0,0} - \sum_{n=2}^N a_n \widehat f(\pi_\lambda)^n_{0,0}= \widehat \Psi(\pi_\lambda)_{0,0} - \sum_{n=2}^N a_n \widehat \Psi(\pi_\lambda)^n_{0,0}. \end{align} $$

We define

$$ \begin{align*} \Omega:= \left\{\lambda \in \mathfrak{a} \mid \widehat f(\pi_\lambda)_{0,0}= \widehat \Psi(\pi_\lambda)_{0,0}\right\}\subset \mathfrak{a}, \end{align*} $$

and by adopting a topological argument, we show that $\Omega = \mathfrak {a}$ . Since both $\lambda \mapsto \widehat f(\pi _\lambda )_{0,0}$ and $\lambda \mapsto \widehat \Psi (\pi _\lambda )_{0,0}$ are continuous, it follows that $\Omega $ is closed in $\mathfrak {a}$ . To show that $\Omega $ is also open in $\mathfrak {a}$ , we first note that by nonnegativity of $\Psi $ and the bound of the spherical functions (2.3), we have for all $\lambda \in \mathfrak {a}$

(4.10) $$ \begin{align} \left|\widehat \Psi(\pi_\lambda)_{0,0}\right| = \left|\int_G \Psi(x)\:\varphi_{\lambda}(x)\:dx\right| \le \int_G \Psi(x)\:\left|\varphi_{\lambda}(x)\right|\:dx \le \int_G \Psi(x)\:dx. \end{align} $$

If $\Psi $ is a nontrivial element of $L^1(G/K)$ , that is,

$$ \begin{align*} E:=\{x \in G \mid \Psi(x)>0\} \end{align*} $$

is a set of positive Haar measure, then we claim that one actually has strict inequality above, that is,

(4.11) $$ \begin{align} \left|\widehat \Psi(\pi_\lambda)_{0,0}\right| < \int_G \Psi(x)\:dx,\:\text{ for all } \lambda \in \mathfrak{a}. \end{align} $$

Indeed, if there exists $\lambda _0 \in \mathfrak {a}$ such that

$$ \begin{align*} \left|\widehat \Psi(\pi_{\lambda_0})_{0,0}\right| = \int_G \Psi(x)\:dx, \end{align*} $$

then one has equality throughout (4.10) for $\lambda _0$ and hence, in particular,

$$ \begin{align*} \int_G \Psi(x)\:\left|\varphi_{\lambda_0}(x)\right|\:dx = \int_G \Psi(x)\:dx. \end{align*} $$

Then using the nonnegativity of $\Psi $ and the definition of E, we get that

$$ \begin{align*} \int_E \Psi(x)\:\left(\left|\varphi_{\lambda_0}(x)\right|-1\right)\:dx = 0. \end{align*} $$

Now positivity of $\Psi $ on E along with the boundedness (2.3) of spherical functions yield the existence of a set of positive Haar measure, say F, such that

$$ \begin{align*} \left|\varphi_{\lambda_0}(x)\right|=1,\:\text{ for all } x \in F. \end{align*} $$

Then the inequalities (2.3) and (2.4) yield for all $x \in F$ ,

$$ \begin{align*} 1 = \left|\varphi_{\lambda_0}(x)\right| \le \varphi_0(x) \le 1, \end{align*} $$

and hence

$$ \begin{align*} \varphi_0 \equiv 1, \text{ on } F. \end{align*} $$

Now as F is a set of positive Haar measure and $\varphi _0$ , being an eigenfunction of $\Delta $ , is real-analytic, we must have that

$$ \begin{align*} \varphi_0 \equiv 1, \text{ on } G, \end{align*} $$

but this contradicts the decay of $\varphi _0$ away from the identity (2.4). Hence, we have the strict inequality (4.11), which combined with (4.2), yields that

$$ \begin{align*} \left|\widehat \Psi(\pi_\lambda)_{0,0}\right| < \int_G \Psi(x)\:dx \le t_{\mathcal Q},\:\text{ for all } \lambda \in \mathfrak{a}. \end{align*} $$

On the other hand, if $\Psi $ is zero almost everywhere, then trivially we have that

$$ \begin{align*} \left|\widehat \Psi(\pi_\lambda)_{0,0}\right| =0 < t_{\mathcal Q},\:\text{ for all } \lambda \in \mathfrak{a}. \end{align*} $$

Hence, in either cases, we have

(4.12) $$ \begin{align} \left|\widehat \Psi(\pi_\lambda)_{0,0}\right| < t_{\mathcal Q},\:\text{ for all } \lambda \in \mathfrak{a}. \end{align} $$

Now given $\lambda _0 \in \Omega $ , there exists $\varepsilon>0$ , such that

(4.13) $$ \begin{align} \left|\widehat \Psi(\pi_{\lambda_0})_{0,0}\right| < (1-\varepsilon)t_{\mathcal Q}. \end{align} $$

By continuity of $\lambda \mapsto \widehat f(\pi _\lambda )_{0,0}$ , there exists $\delta>0$ such that

(4.14) $$ \begin{align} \left|\left|\hspace{1.5pt} \widehat f(\pi_{\lambda_0})_{0,0}\right|-\left|\hspace{1.5pt} \widehat f(\pi_\lambda)_{0,0}\right|\right| < \varepsilon\: t_{\mathcal Q},\:\text{ for all } \lambda \in B(\lambda_0,\delta), \end{align} $$

where $B(\lambda _0,\delta )$ is the ball in $\mathfrak {a}$ with center $\lambda _0$ and radius $\delta $ with respect to the norm on $\mathfrak {a}$ inherited from $\mathfrak {p}$ . Now as $\lambda _0 \in \Omega $ , by (4.13), we have

$$ \begin{align*} \left|\hspace{1.5pt}\widehat f(\pi_{\lambda_0})_{0,0}\right|=\left|\hspace{1.5pt}\widehat \Psi(\pi_{\lambda_0})_{0,0}\right| < (1-\varepsilon)t_{\mathcal Q}, \end{align*} $$

which combined with (4.14) yields

$$ \begin{align*} \left|\hspace{1.5pt}\widehat f(\pi_\lambda)_{0,0}\right| < t_{\mathcal Q}\:,\:\text{ for all } \lambda \in B(\lambda_0,\delta). \end{align*} $$

So, in particular, both

$$ \begin{align*} \left|\widehat \Psi(\pi_\lambda)_{0,0}\right|<t_{\mathcal Q}\:\text{ and }\:\left|\hspace{1.5pt}\widehat f(\pi_\lambda)_{0,0}\right| <t_{\mathcal Q},\:\text{ for all } \lambda \in B(\lambda_0,\delta). \end{align*} $$

Then in view of (4.9) and the fact that the complex polynomial $\mathcal Q$ is injective on the disk $\mathcal {D}$ in ${\mathbb {C}}$ with center $0$ and radius $t_{\mathcal Q}$ (whose proof is contained in the proof of Theorem 1.2), it follows that

$$ \begin{align*} \widehat \Psi(\pi_\lambda)_{0,0} =\widehat f(\pi_\lambda)_{0,0},\:\text{ for all } \lambda \in B(\lambda_0,\delta). \end{align*} $$

Thus, $B(\lambda _0,\delta ) \subset \Omega $ and hence $\Omega $ is also open in $\mathfrak {a}$ .

We finally show that $\Omega $ is non-empty. To show this, we consider $f^\#$ , the left K-average of f,

$$ \begin{align*} f^\#(g):=\int_K f(kg)\:dk,\text{ for } g \in G. \end{align*} $$

Now as $f \in L^1(G/K)$ , we note that $f^\# \in L^1(G//K)$ and moreover, for all $\lambda \in \mathfrak {a}$ ,

$$ \begin{align*} \widehat f(\pi_{\lambda})_{0,0} = \mathcal H f^\#(\lambda), \end{align*} $$

the spherical Fourier transform of $f^\#$ . Then, by the Riemann–Lebesgue lemma (2.5) on $\mathcal H f^\#$ , there exists $\lambda _0 \in \mathfrak {a}$ such that

$$ \begin{align*} \left|\hspace{1.5pt}\widehat f(\pi_{\lambda_0})_{0,0}\right| < t_{\mathcal Q}. \end{align*} $$

Thus, by (4.12), we get that both $\widehat f(\pi _{\lambda _0})_{0,0},\: \widehat \Psi (\pi _{\lambda _0})_{0,0} \in \mathcal {D}$ and then by (4.9) and the injectivity of $\mathcal Q$ on $\mathcal {D}$ , we obtain $\widehat f(\pi _{\lambda _0})_{0,0}\:=\: \widehat \Psi (\pi _{\lambda _0})_{0,0}$ , that is, $\lambda _0 \in \Omega $ and hence $\Omega $ is non-empty. Since $\Omega $ is a non-empty subset of $\mathfrak {a}$ which is both closed as well as open, we have $\Omega =\mathfrak {a}$ , that is,

(4.15) $$ \begin{align} \widehat f(\pi_\lambda)_{0,0}\:=\: \widehat \Psi(\pi_\lambda)_{0,0},\:\text{ for all } \lambda \in \mathfrak{a}. \end{align} $$

Now in view of (4.12), we reconsider the complex polynomial

$$ \begin{align*} \mathcal Q(z) = z - \sum_{n=2}^N a_n z^n = z - \mathcal P(z), \end{align*} $$ script upper Q left parenthesis z right parenthesis equals z minus sigma summation Underscript n equals 2 Overscript upper N Endscripts a Subscript n Baseline z Superscript n Baseline equals z minus script upper P left parenthesis z right parenthesis comma

and now aim to show that the polynomial

$$ \begin{align*} \mathcal Q_1(z):=\frac{\mathcal Q(z)}{z}=1-\frac{\mathcal P(z)}{z} \end{align*} $$

has no zeros in $\mathcal {D}$ . To show this, it suffices to prove that

(4.16) $$ \begin{align} \left|\frac{\mathcal P(z)}{z}\right|<1,\:\text{ for all } z \in \mathcal{D}. \end{align} $$ StartAbsoluteValue StartFraction script upper P left parenthesis z right parenthesis Over z EndFraction EndAbsoluteValue less than 1 comma for all z element of script upper D period

This is clearly true for $z=0$ . To show (4.16) for $z \in \mathcal {D}\setminus \{0\}$ , we first note that $\mathcal P(0)=0$ and then recall (3.21) which states that

$$ \begin{align*} \left|\mathcal P'(z)\right| < 1,\text{ for all } z \in \mathcal{D}. \end{align*} $$

Then for any $z \in \mathcal {D}\setminus \{0\}$ , we get

$$ \begin{align*} \left|\mathcal P(z)\right| = \left|\mathcal P(z)-\mathcal P(0)\right| =\left|\int_0^z \mathcal P'(\xi)\:d\xi\right|<|z|. \end{align*} $$

Thus, we get (4.16) as

$$ \begin{align*} \left|\frac{\mathcal P(z)}{z}\right|=\frac{\left|\mathcal P(z)\right|}{\left|z\right|}<1. \end{align*} $$ StartLayout 1st Row StartAbsoluteValue StartFraction script upper P left parenthesis z right parenthesis Over z EndFraction EndAbsoluteValue equals StartFraction StartAbsoluteValue script upper P left parenthesis z right parenthesis EndAbsoluteValue Over StartAbsoluteValue z EndAbsoluteValue EndFraction less than 1 period EndLayout

Hence, $\mathcal Q_1$ has no zeros in $\mathcal {D}$ . Now by plugging (4.15) in (4.8), we note that

$$ \begin{align*} \widehat f(\pi_\lambda)_{p,0}\: \mathcal Q_1\left(\widehat \Psi(\pi_\lambda)_{0,0}\right)=\widehat \Psi(\pi_\lambda)_{p,0}\: \mathcal Q_1\left(\widehat \Psi(\pi_\lambda)_{0,0}\right). \end{align*} $$

Then in view of (4.12) and the fact that $\mathcal Q_1$ has no zeros in $\mathcal {D}$ , we get that

$$ \begin{align*} \widehat f(\pi_\lambda)_{p,0}\:=\: \widehat \Psi(\pi_\lambda)_{p,0},\:\text{ for all } \lambda \in \mathfrak{a}\:, p \in \mathbb N \cup \{0\}. \end{align*} $$

The above along with (4.6) yields (4.3) which completes the proof of Theorem 1.4.

We now study the K-bi-invariant extremizers of the convolution inequality $f \ge f*f$ on G. But before that, we recall its Euclidean counterpart.

Lemma 4.1 [Reference Carlen, Jauslin, Lieb and LossCJLL21, Theorem 2]

Let $f \in L^1(\mathbb R^d)$ satisfy $f \ge f*f$ and $\int _{\mathbb R^d} f=1/2$ . Then $\int _{\mathbb R^d} |x|\:f(x)\:dx=\infty $ .

We now present the proof of Theorem 1.5.

Proof of Theorem 1.5

In this proof, we will make use of the convolutions on both G, as well as $\mathfrak {a}$ (as the additive group $\mathbb R^d$ , where $d=rank(G)$ ) and hence will mention the group under consideration to avoid confusion. Let $f \in L^1(G//K)$ be as in the statement of Theorem 1.5, that is, f satisfies

(4.17) $$ \begin{align} & f(x) \ge f*_G f(x),\text{ almost every } x \in G, \end{align} $$

(4.18) $$ \begin{align} & \int_G f(x)\:dx=\frac{1}{2}. \end{align} $$

We now consider the Abel transform defined in (2.6). Then $\mathscr {A}f \in L^1(\mathfrak {a})^W$ . Moreover, it follows from (4.17) that

(4.19) $$ \begin{align} \mathscr{A}f \ge \mathscr{A}\left(f*_G f\right). \end{align} $$

Now by repeated applications of (2.7), we have that for $\lambda \in \mathfrak {a}$ ,

$$ \begin{align*} \mathscr F\left(\mathscr{A}\left(f*_G f\right)\right)(\lambda)=\left(f*_G f\right)^\wedge(\lambda)=\widehat{f}(\lambda)^2=\mathscr F\left(\mathscr{A}f\right)(\lambda)^2= \mathscr F\left(\mathscr{A}f*_{\mathfrak{a}}\mathscr{A}f\right)(\lambda), \end{align*} $$

and hence,

$$ \begin{align*} \mathscr{A}\left(f*_G f\right) = \mathscr{A}f*_{\mathfrak{a}}\mathscr{A}f, \end{align*} $$

which along with (4.19) yields

(4.20) $$ \begin{align} \mathscr{A}f \ge \mathscr{A}f*_{\mathfrak{a}}\mathscr{A}f. \end{align} $$

We set $\gamma $ on $\mathfrak {a}$ (almost everywhere) by

$$ \begin{align*} \gamma(H):= e^{\rho(H)}\mathscr{A}f(H). \end{align*} $$

By Theorem 1.4, $f \ge 0$ and hence so is $\gamma $ . Then combining (4.18), the Iwasawa decomposition of the Haar measure given by (2.1), the left K-invariance of f, the Fubini–Tonelli theorem, and the definitions of the Abel transform and $\gamma $ , it follows that

(4.21) $$ \begin{align} \gamma \ge 0,\: \gamma \in L^1(\mathfrak{a}) \: \text{ with } \int_{\mathfrak{a}} \gamma(H)\:dH = \frac{1}{2}. \end{align} $$

Indeed,

$$ \begin{align*} \frac{1}{2} = \int_G f(x)\:dx &= \int_K \int_{\mathfrak{a}} \int_N f\left(k\exp\left(H\right) n\right)\:e^{2\rho(H)}\:dk\:dH\:dn \\ &= \int_{\mathfrak{a}} \int_N f\left(\exp\left(H\right) n\right)\:e^{2\rho(H)}\:dH\:dn \\ &= \int_{\mathfrak{a}} e^{\rho(H)}\mathscr{A}f(H)\:dH \\ &= \int_{\mathfrak{a}} \gamma(H)\:dH. \end{align*} $$

StartLayout 1st Row 1st Column one half equals integral Underscript upper G Endscripts f left parenthesis x right parenthesis d x 2nd Column equals integral Underscript upper K Endscripts integral Underscript German a Endscripts integral Underscript upper N Endscripts f left parenthesis k exp left parenthesis upper H right parenthesis n right parenthesis e Superscript 2 rho left parenthesis upper H right parenthesis Baseline d k d upper H d n 2nd Row 1st Column Blank 2nd Column equals integral Underscript German a Endscripts integral Underscript upper N Endscripts f left parenthesis exp left parenthesis upper H right parenthesis n right parenthesis e Superscript 2 rho left parenthesis upper H right parenthesis Baseline d upper H d n 3rd Row 1st Column Blank 2nd Column equals integral Underscript German a Endscripts e Superscript rho left parenthesis upper H right parenthesis Baseline script upper A f left parenthesis upper H right parenthesis d upper H 4th Row 1st Column Blank 2nd Column equals integral Underscript German a Endscripts gamma left parenthesis upper H right parenthesis d upper H period EndLayout

Moreover, using (4.20), we have for almost every $H \in \mathfrak {a}$

(4.22) $$ \begin{align} \left(\gamma *_{\mathfrak{a}} \gamma\right)(H) &= \int_{\mathfrak{a}} \gamma(H-H')\gamma(H')\:dH' \nonumber\\ &= \int_{\mathfrak{a}} e^{\rho(H-H')}\mathscr{A}f(H-H')\:e^{\rho(H')}\mathscr{A}f(H')\:dH' \nonumber\\ &= e^{\rho(H)} \int_{\mathfrak{a}} \mathscr{A}f(H-H')\:\mathscr{A}f(H')\:dH' \nonumber\\ &= e^{\rho(H)} \left(\mathscr{A}f*_{\mathfrak{a}}\mathscr{A}f\right)(H) \nonumber\\ & \le e^{\rho(H)}\mathscr{A}f(H) \nonumber\\ &= \gamma(H). \end{align} $$

StartLayout 1st Row 1st Column left parenthesis gamma asterisk Subscript German a Baseline gamma right parenthesis left parenthesis upper H right parenthesis 2nd Column equals integral Underscript German a Endscripts gamma left parenthesis upper H minus upper H Superscript prime Baseline right parenthesis gamma left parenthesis upper H Superscript prime Baseline right parenthesis d upper H Superscript prime Baseline 2nd Row 1st Column Blank 2nd Column equals integral Underscript German a Endscripts e Superscript rho left parenthesis upper H minus upper H Super Superscript prime Superscript right parenthesis Baseline script upper A f left parenthesis upper H minus upper H Superscript prime Baseline right parenthesis e Superscript rho left parenthesis upper H Super Superscript prime Superscript right parenthesis Baseline script upper A f left parenthesis upper H Superscript prime Baseline right parenthesis d upper H Superscript prime Baseline 3rd Row 1st Column Blank 2nd Column equals e Superscript rho left parenthesis upper H right parenthesis Baseline integral Underscript German a Endscripts script upper A f left parenthesis upper H minus upper H Superscript prime Baseline right parenthesis script upper A f left parenthesis upper H Superscript prime Baseline right parenthesis d upper H Superscript prime Baseline 4th Row 1st Column Blank 2nd Column equals e Superscript rho left parenthesis upper H right parenthesis Baseline left parenthesis script upper A f asterisk Subscript German a Baseline script upper A f right parenthesis left parenthesis upper H right parenthesis 5th Row 1st Column Blank 2nd Column less than or equals e Superscript rho left parenthesis upper H right parenthesis Baseline script upper A f left parenthesis upper H right parenthesis 6th Row 1st Column Blank 2nd Column equals gamma left parenthesis upper H right parenthesis period EndLayout

Then combining (4.21), (4.22), and Lemma 4.1, it follows that

$$ \begin{align*} \int_{\mathfrak{a}} \left|H\right|\:\gamma(H)\:dH=\infty. \end{align*} $$

Then plugging in the definitions of $\gamma $ , the Abel transform, the Fubini–Tonelli theorem, the Iwasawa decomposition of the Haar measure, and the left K-invariance of f, the last equation becomes

$$ \begin{align*} \int_G |H(x)|\:f(x)\:dx=\infty. \end{align*} $$

Finally, using the fact that $f \ge 0$ and invoking the relation (2.2), we obtain

$$ \begin{align*} \int_G \|x\|\:f(x)\:dx \ge \int_G |H(x)|\:f(x)\:dx=\infty.\\[-34pt] \end{align*} $$

Next, we prove Theorem 1.7.

Proof of Theorem 1.7

For $\xi>0$ , the resolvent operator $\left (\xi I - \mathcal L\right )^{-1}$ is realized by convolution on the right with the K-bi-invariant tempered distribution $K_\xi $ on G given by

$$ \begin{align*} K_\xi(x)= \int_0^\infty e^{-t\xi}\:h_t(x)\:dt, \end{align*} $$

where $h_t$ is the heat kernel on $\mathbb X$ . Hence, $K_\xi $ is nonnegative and moreover, an application of the Fubini–Tonelli theorem yields

(4.23) $$ \begin{align} \int_G K_\xi(x) dx = \int_G \int_0^\infty e^{-t\xi}\:h_t(x)\:dt\:dx = \int_0^\infty e^{-t\xi}\:dt= \frac{1}{\xi}. \end{align} $$

Now the integro-differential equation (1.9) can be rewritten as

$$ \begin{align*} u = {\left(-\mathcal L +\xi \right)}^{-m}\left\{V(1-u)\right\}+{\left(-\mathcal L +\xi \right)}^{-m}\left\{\mu \left(*^{m+1} u\right)\right\}. \end{align*} $$

Then as $K_\xi ,V,$ and u all are K-bi-invariant, the convolutions commute and we can write

$$ \begin{align*} u=\left(*^m K_\xi\right)*\left\{V(1-u)\right\} + \mu \left\{*^m \left(K_\xi * u\right)\right\}*u. \end{align*} $$

Now defining

$$ \begin{align*} f:= \mu^{1/m} \left(K_\xi *u\right), \end{align*} $$

we note that $f \in L^1(G//K)$ is real-valued and then rewrite the above as

(4.24) $$ \begin{align} u=\left(*^m K_\xi\right)*\left\{V(1-u)\right\} + \left(*^m f\right)*u. \end{align} $$

Next, by integrating out f and plugging in the constraint (1.10) and (4.23), we obtain

(4.25) $$ \begin{align} \int_G f = \mu^{1/m}\left(\int_G K_\xi\right)\left(\int_G u\right)=\frac{\mu^{1/m}}{\xi \delta}, \end{align} $$ StartLayout 1st Row integral Underscript upper G Endscripts f equals mu Superscript 1 divided by m Baseline left parenthesis integral Underscript upper G Endscripts upper K Subscript xi Baseline right parenthesis left parenthesis integral Underscript upper G Endscripts u right parenthesis equals StartFraction mu Superscript 1 divided by m Baseline Over xi delta EndFraction comma EndLayout

and thus in view of the hypothesis (1.11), it follows that

$$ \begin{align*} \int_G f \in (0,1). \end{align*} $$

Now convolving (4.24) with $\mu ^{1/m} K_\xi $ , we get

$$ \begin{align*} f= \mu^{1/m}\left(*^{m+1} K_\xi\right)*\left\{V(1-u)\right\} + \left(*^{m+1} f\right). \end{align*} $$

Then as $K_\xi $ is nonnegative and from the hypothesis, we also have that V is nonnegative and $u \le 1$ almost everywhere on G, it follows that

$$ \begin{align*} f \ge *^{m+1}f. \end{align*} $$

Then in view of (4.25), we note that Theorem 1.4 is applicable on f to conclude that f is nonnegative. Now define for any real-valued h,

$$ \begin{align*} h_{-}(x):=\begin{cases} -h(x),\: &\text{ if } h(x)< 0, \\ 0,\: &\text{ if } h(x) \ge 0, \end{cases} \end{align*} $$ StartLayout 1st Row h Subscript minus Baseline left parenthesis x right parenthesis colon equals StartLayout Enlarged left brace 1st Row 1st Column minus h left parenthesis x right parenthesis comma 2nd Column if h left parenthesis x right parenthesis less than 0 comma 2nd Row 1st Column 0 comma 2nd Column if h left parenthesis x right parenthesis greater than or equals 0 comma EndLayout EndLayout

and also

$$ \begin{align*} h_{+}(x):=\begin{cases} 0,\: &\text{ if } h(x)< 0, \\ h(x),\: &\text{ if } h(x) \ge 0, \end{cases} \end{align*} $$ StartLayout 1st Row h Subscript plus Baseline left parenthesis x right parenthesis colon equals StartLayout Enlarged left brace 1st Row 1st Column 0 comma 2nd Column if h left parenthesis x right parenthesis less than 0 comma 2nd Row 1st Column h left parenthesis x right parenthesis comma 2nd Column if h left parenthesis x right parenthesis greater than or equals 0 comma EndLayout EndLayout

and thus $h=h_+-h_-$ . Then by (4.24) and nonnegativity of $K_\xi $ , V, $1-u,$ and f, it follows that

$$ \begin{align*} u_{-} &= \left[\left(*^m K_\xi\right)*\left\{V(1-u)\right\} + \left(*^m f\right)*u\right]_{-} \\ & \le \left[ \left(*^m f\right)*u\right]_{-} \\ &= \left[ \left(*^m f\right)*u_+ - \left(*^m f\right)*u_-\right]_{-} \\ & \le \left(*^m f\right)*u_-. \end{align*} $$

StartLayout 1st Row 1st Column u Subscript minus 2nd Column equals left bracket left parenthesis asterisk Superscript m Baseline upper K Subscript xi Baseline right parenthesis asterisk StartSet upper V left parenthesis 1 minus u right parenthesis EndSet plus left parenthesis asterisk Superscript m Baseline f right parenthesis asterisk u right bracket Subscript minus Baseline 2nd Row 1st Column Blank 2nd Column less than or equals left bracket left parenthesis asterisk Superscript m Baseline f right parenthesis asterisk u right bracket Subscript minus Baseline 3rd Row 1st Column Blank 2nd Column equals left bracket left parenthesis asterisk Superscript m Baseline f right parenthesis asterisk u Subscript plus Baseline minus left parenthesis asterisk Superscript m Baseline f right parenthesis asterisk u Subscript minus Baseline right bracket Subscript minus Baseline 4th Row 1st Column Blank 2nd Column less than or equals left parenthesis asterisk Superscript m Baseline f right parenthesis asterisk u Subscript minus Baseline period EndLayout

Integrating the above inequality and then invoking (4.25) yields

$$ \begin{align*} \int_G u_- \le {\left(\int_G f \right)}^m \left(\int_G u_- \right) = \frac{\mu}{\xi^m \delta^m} \left(\int_G u_- \right). \end{align*} $$

The above inequality combined with the hypothesis (1.11) yields

$$ \begin{align*} \int_G u_- = 0, \end{align*} $$

which in turn implies that $u_- = 0$ almost everywhere on G and hence $u \ge 0$ almost everywhere on G. This completes the proof of Theorem 1.7.

Acknowledgements

The author would like to thank the anonymous referee for helpful suggestions which improved the exposition. The author is thankful to Prof. Swagato K. Ray for bringing [Reference Carlen, Jauslin, Lieb and LossCJLL21] to his attention.

Footnotes

The author is supported by the Institute Post-Doctoral Fellowship of Indian Institute of Technology Bombay.

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