Generalized Bessel and Frame Measures

TL;DR

This paper introduces and explores generalized notions of Bessel and frame measures in $L^p$ spaces with respect to a measure, providing new constructions, properties, and examples of such measures.

Contribution

It defines $(p,q)$-Bessel and frame measures, extends classical concepts, and offers methods to construct and analyze these measures in a broad setting.

Findings

Every finite Borel measure is a $(p,q)$-Bessel measure for some measure.

Many measures admit infinite $(p,q)$-Bessel measures.

$(p,q)$-Bessel/frame measures are $\sigma$-finite and non-unique.

Abstract

Considering a finite Borel measure $ \mu $ on $ \mathbb{R}^d $, a pair of conjugate exponents $ p, q $, and a compatible semi-inner product on $ L^p(\mu) $, we introduce $ (p,q) $-Bessel and $ (p,q) $-frame measures as a generalization of the concepts of Bessel and frame measures. In addition, we define notions of $ q $-Bessel and $ q$-frame in the semi-inner product space $ L^p(\mu) $. Every finite Borel measure $\nu$ is a $(p,q)$-Bessel measure for a finite measure $ \mu $. We construct a large number of examples of finite measures $ \mu $ which admit infinite $ (p,q) $-Bessel measures $ \nu $. We show that if $ \nu $ is a $ (p,q) $-Bessel/frame measure for $ \mu $, then $ \nu $ is $ \sigma $-finite and it is not unique. In fact, by using convolutions of probability measures, one can obtain other $ (p,q) $-Bessel/frame measures for $ \mu $. We present a general way of constructing a $ (p,q) $-Bessel/frame measure for a given measure.

Equations

$${\int }_{{\mathbb{R}}^d}|\widehat{fd\mu }(t){|}^{2}d\nu (t)\simeq {\int }_{{\mathbb{R}}^d}|f(x){|}^{2}d\mu (x),$$

$$\widehat{fd\mu }(t)={\int }_{{\mathbb{R}}^d}f(x)e^{-2\pi it\cdot x}d\mu (x)(t\in {\mathbb{R}}^d).$$

$$A{\int }_{{\mathbb{R}}^d}|f(x){|}^{2}d\mu (x)\le {\int }_{{\mathbb{R}}^d}|\widehat{fd\mu }(t){|}^{2}d\nu (t)\le B{\int }_{{\mathbb{R}}^d}|f(x){|}^{2}d\mu (x).$$

$$e_t(x)=e^{2\pi it\cdot x}(x\in {\mathbb{R}}^d).$$

$$\sum _{i\in I}|⟨f,f_i⟩{|}^2\le B\Vert f{\Vert }^2.$$

$$A\Vert f{\Vert }^2\le \sum _{i\in I}|⟨f,f_i⟩{|}^2\le B\Vert f{\Vert }^2.$$

$$A\Vert f{\Vert }_{L^2(\mu )}^2\le \sum _{\lambda \in \Lambda }|{⟨f,e_{\lambda }⟩}_{L^2(\mu )}{|}^2\le B\Vert f{\Vert }_{L^2(\mu )}^2.$$

$$\Vert \widehat{fd\mu }{\Vert }_{L^2(\nu )}^2\le B\Vert f{\Vert }_{L^2(\mu )}^2.$$

$$A\Vert f{\Vert }_{L^2(\mu )}^2\le \Vert \widehat{fd\mu }{\Vert }_{L^2(\nu )}^2\le B\Vert f{\Vert }_{L^2(\mu )}^2.$$

$${\tau }_a(x)=R^{-1}(x+a)(x\in {\mathbb{R}}^d,a\in \mathcal{A}).$$

$$[f,g{]}_{L^p(\mu )}:=\frac{1}{\Vert g{\Vert }_{L^p(\mu )}^{p-2}}{\int }_{X}f(x)|g(x){|}^{p-1}\overline{sgn(g(x))}d\mu (x),$$

$$A\Vert f{\Vert }_X\le \Vert \{[f,f_i]{\}}_{i\in I}{\Vert }_{X_d}\le B\Vert f{\Vert }_X.$$

$$\sum _{i\in I}|[f,f_i{]}_{L^p(\mu )}{|}^q\le B\Vert f{\Vert }_{L^p(\mu )}^q.$$

$$A\Vert f{\Vert }_{L^p(\mu )}^q\le \sum _{i\in I}|[f,f_i{]}_{L^p(\mu )}{|}^q\le B\Vert f{\Vert }_{L^p(\mu )}^q.$$

$${\int }_{{\mathbb{R}}^d}|[f,e_t{]}_{L^p(\mu )}{|}^{q}d\nu (t)\le B\Vert f{\Vert }_{L^p(\mu )}^q(p\ne 1,q\ne \infty )$$

$$\Vert \widehat{fd\mu }{\Vert }_{\infty }\le B\Vert f{\Vert }_{L^1(\mu )}(p=1,q=\infty ).$$

$$A\Vert f{\Vert }_{L^p(\mu )}^q\le {\int }_{{\mathbb{R}}^d}|[f,e_t{]}_{L^p(\mu )}{|}^{q}d\nu (t)\le B\Vert f{\Vert }_{L^p(\mu )}^q(p\ne 1,q\ne \infty )$$

$$A\Vert f{\Vert }_{L^1(\mu )}\le \Vert \widehat{fd\mu }{\Vert }_{\infty }\le B\Vert f{\Vert }_{L^1(\mu )}(p=1,q=\infty ).$$

$$|[f,e_t{]}_{L^p(\mu )}|\le {\int }_{{\mathbb{R}}^d}|f(x)e^{-2\pi it\cdot x}|d\mu (x)\le (\mu ({\mathbb{R}}^d){)}^{\frac{1}{q}}\Vert f{\Vert }_{L^p(\mu )}.$$

$${\int }_{{\mathbb{R}}^d}|[f,e_t{]}_{L^p(\mu )}{|}^{q}d\nu (t)\le \mu ({\mathbb{R}}^d)\nu ({\mathbb{R}}^d)\Vert f{\Vert }_{L^p(\mu )}^q.$$

$$\sum _{\lambda \in \Lambda }|[f,e_{\lambda }{]}_{L^p(\mu )}{|}^q={\int }_{{\mathbb{R}}^d}|[f,e_t{]}_{L^p(\mu )}{|}^{q}d\nu (t),$$

$$\sum _{\lambda \in \Lambda }|[f,e_{\lambda }{]}_{L^p(\mu )}{|}^q={\int }_{{\mathbb{R}}^d}|[f,e_t{]}_{L^p(\mu )}{|}^{q}d\nu (t).$$

$${\int }_{{\mathbb{R}}^d}|[f,e_t{]}_{L^p(\mu )}{|}^{q}d\nu (t)=\sum _{\lambda \in \Lambda }{d}_{\lambda }|[f,e_{\lambda }{]}_{L^p(\mu )}{|}^q=\sum _{\lambda \in \Lambda }|[f,\sqrt[q]{d_{\lambda }}{e}_{\lambda }]{|}^q.$$

$${\int }_{[0,1{]}^d}|f(x){|}^p{\left|\frac{{\varphi }^{p-1}(x)}{\Vert \varphi {\Vert }_p^{p-2}}\right|}^{p}d(x)\le \frac{b^{(p-1)p}}{a^{(p-2)p}}{\int }_{[0,1{]}^d}|f(x){|}^{p}d(x)<\infty .$$

$$\sum _{t\in {\mathbb{Z}}^d}{\left|[f,{\varphi }_t{e}_t{]}_{L^p([0,1{]}^d)}\right|}^q$$

$$\le {\left|{\int }_{[0,1{]}^d}|f(x){|}^p{\left|\frac{{\varphi }^{p-1}(x)}{\Vert \varphi {\Vert }_p^{p-2}}\right|}^{p}dx\right|}^{q/p}\le \frac{b^p}{a^{p-q}}\Vert f{\Vert }_p^q.$$

$$\frac{a^p}{b^{p-q}}A\parallel f{\parallel }_{L^p(\mu )}^q\le \Vert \{[f,{\varphi }_i{f}_i{]}_{L^p(\mu )}{\}}_{i\in I}{\Vert }^q\le \frac{b^p}{a^{p-q}}B\Vert f{\Vert }_{L^p(\mu )}^q.$$

$$\frac{{\partial }^k\widehat{fd\mu }}{\partial t_1^k}(t)=\int f(x)(-2\pi i{x}_1{)}^k{e}^{-2\pi it\cdot x}d\mu (x)={\left[(-2\pi i{x}_1{)}^{k}f,e_t\right]}_{L^p(\mu )},(t\in {\mathbb{R}}^d).$$

$$|\widehat{fd\mu }({\omega }_n)-\widehat{fd\mu }({\lambda }_n){|}^q$$

$$\le \sum _{k=1}^{\infty }\frac{{\left|\frac{{\partial }^k\widehat{fd\mu }}{\partial t_1^k}({\lambda }_n)\right|}^q}{k!}\cdot {\left(\sum _{k=1}^{\infty }\frac{|({\omega }_n{)}_1-({\lambda }_n{)}_1{|}^{pk}}{k!}\right)}^{q/p}$$

$$\le \sum _{k=1}^{\infty }\frac{{\left|\frac{{\partial }^k\widehat{fd\mu }}{\partial t_1^k}({\lambda }_n)\right|}^q}{k!}\cdot {\left(\sum _{k=1}^{\infty }\frac{C^{pk}}{k!}\right)}^{q-1}=\sum _{k=1}^{\infty }\frac{{\left|\frac{{\partial }^k\widehat{fd\mu }}{\partial t_1^k}({\lambda }_n)\right|}^q}{k!}\cdot {\left(e^{C^p}-1\right)}^{q-1}.$$

$$\sum _{n\in \mathbb{N}}{\left|\frac{{\partial }^k\widehat{fd\mu }}{\partial t_1^k}({\lambda }_n)\right|}^q=\sum _{n\in \mathbb{N}}{\left|{\left[(-2\pi i{x}_1{)}^{k}f,e_{{\lambda }_n}\right]}_{L^p(\mu )}\right|}^q\le B\Vert (-2\pi i{x}_1{)}^{k}f{\Vert }_{L^p(\mu )}^q\le B(2\pi M{)}^{qk}\Vert f{\Vert }_{L^p(\mu )}^q.$$

Full text

Keywords: Fourier frame, Plancherel theorem, spectral measure, frame measure, Bessel measure, semi-inner product.

2010 AMS Subject Classification: Primary 28A99, 46E30, 42C15.

1 Introduction

According to [5], a Borel measure $\nu$ on $\mathbb{R}^{d}$ is called a dual measure for a given measure $\mu$ on $\mathbb{R}^{d}$ if for every $f\in L^{2}(\mu)$,

[TABLE]

where for a function $f\in L^{1}(\mu)$ the Fourier transform is given by

[TABLE]

Precisely, the equivalence in Equation (1) means that there are positive constants $A$ and $B$ independent of the function $f(x)$ such that

[TABLE]

Therefore when $A=B=1$, by Plancherel’s theorem for Lebesgue measure $\lambda$ on $\mathbb{R}^{d}$, $\lambda$ is a dual measure to itself. Dual measures are in fact a generalization of the concept of Fourier frames and they are also called frame measures. According to [5], if $\mu$ is not an $F$-spectral measure, then there cannot be any general statement about the existence of frame measures $\nu$. Nevertheless, the authors showed that if one frame measure exists, then by using convolutions of measures, many frame measures can be obtained, especially a frame measure which is absolutely continuous with respect to Lebesgue measure. Moreover, they presented a general way of constructing Bessel/frame measures for a given measure.

In this paper we generalize the notion of Bessel/frame measure from Hilbert spaces $L^{2}(\mu)$, $L^{2}(\nu)$ to Banach spaces $L^{p}(\mu)$, $L^{q}(\nu)$ ($p,q$ are conjugate exponents) via a compatible semi-inner product defined on $L^{p}(\mu)$. Compatible semi-inner products are natural substitutes for inner products on Hilbert spaces. We introduce $(p,q)$-Bessel and $(p,q)$-frame measures, and we define notions of $q$-Bessel and $q$-frame in the semi-inner product space $L^{p}(\mu)$. Then we investigate the existence and some general properties of them.

The rest of this paper is organized as follows: In section $2$ basic definitions and preliminaries are given. In section $3$ we investigate the existence of $(p,q)$-Bessel/frame measures. We show that every finite Borel measure $\nu$ is a $(p,q)$-Bessel measure for a finite measure $\mu$. In addition, we construct a large number of examples of measures which admit infinite discrete $(p,q)$-Bessel measures, by F-spectral measures and applying the Riesz-Thorin interpolation theorem. In general, for every spectral measure (B-spectral measure, or F-spectral measure respectively) $\mu$, there exists a discrete measure $\nu=\sum_{\lambda\in\Lambda_{\mu}}\delta_{\lambda}$ which is a Plancherel measure (Bessel measure or frame measure respectively) for $\mu$. Then the Riesz-Thorin interpolation theorem yields that $\nu$ is also a $(p,q)$-Bessel measure for $\mu$, where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$. Moreover, this shows that if $\mu$ is a spectral measure (B-spectral measure, or F-spectral measure), then the set $\{e_{\lambda}\}_{\lambda\in\Lambda_{\mu}}$ forms a $q$-Bessel for $L^{p}(\mu)$. It is known [13, 19] that if a measure $\mu$ is an F-spectral measure, then it must be of pure type, i.e., $\mu$ is either discrete, absolutely continuous or singular continuous. Therefore, we consider such measures in constructing the examples. The interested reader can refer to [3, 6, 7, 9, 13, 16, 18, 19, 20, 21, 23, 24] to see examples and properties of spectral measures (B-spectral measures, or F-spectral measures) and related concepts. Besides discrete $(p,q)$-Bessel measures $\nu=\sum_{\lambda\in\Lambda_{\mu}}\delta_{\lambda}$ associated to spectral measures (B-spectral measures, or F-spectral measures) $\mu$, we prove that there exists an infinite absolutely continuous $(p,q)$-Bessel measure $\nu$ for a special finite measure $\mu$. We show that if $\nu$ is a $(p_{1},q_{1})$-Bessel/frame measure and $(p_{2},q_{2})$-Bessel/frame measure for $\mu$, where $1\leq p_{1},p_{2}<\infty$ and $q_{1},q_{2}$ are the conjugate exponents to $p_{1},p_{2}$, respectively, then $\nu$ is a $(p,q)$-Bessel measure for $\mu$ too, where $p_{1}<p<p_{2}$ and $q$ is the conjugate exponent to $p$. Consequently, if $\nu$ is a Bessel/frame measure for $\mu$, then it is a $(p,q)$-Bessel measure for $\mu$ too. In Proposition 3.30 we prove that there exists a measure $\mu$ which admits tight $(p,q)$-frame measures and $(p,q)$-Plancherel measures.

Section $4$ is devoted to investigating properties of $(p,q)$-Bessel/frame measures based on the results by Dutkay, Han, and Weber from [5].

2 Preliminaries

Definition 2.1**.**

Let $t\in\mathbb{R}^{d}$. Denote by $e_{t}$ the exponential function

[TABLE]

Definition 2.2**.**

Let $H$ be a Hilbert space. A sequence $\{f_{i}\}_{i\in I}$ of elements in $H$ is called a Bessel sequence for $H$ if there exists a positive constant $B$ such that for all $f\in H$,

[TABLE]

Here $B$ is called the Bessel bound for the Bessel sequence $\{f_{i}\}_{i\in I}$.

The sequence $\{f_{i}\}_{i\in I}$ is called a frame for $H$, if there exist constants $A,B>0$ such that for all $f\in H$,

[TABLE]

In this case, $A$ and $B$ are called frame bounds.

Frames are a natural generalization of orthonormal bases. It is easily seen from the lower bound that a frame is complete in H, so every $f$ can be expressed using (infinite) linear combination of the elements $f_{i}$ in the frame [2].

Definition 2.3**.**

Let $\mu$ be a compactly supported probability measure on $\mathbb{R}^{d}$ and $\Lambda$ be a countable set in $\mathbb{R}^{d}$, the set $E(\Lambda)=\{e_{\lambda}:\lambda\in\Lambda\}$ is called a Fourier frame for $L^{2}(\mu)$ if for all $f\in L^{2}(\mu)$,

[TABLE]

When $E(\Lambda)$ is an orthonormal basis (Bessel sequence, or frame) for $L^{2}(\mu)$, we say that $\mu$ is a spectral measure (B-spectral measure, or F-spectral measure respectively) and $\Lambda$ is called a spectrum (B-spectrum, or F-spectrum respectively) for $\mu$.

We give the following definition from [5], assuming that the given measure $\mu$ is a finite Borel measure on $\mathbb{R}^{d}$.

Definition 2.4** ([5]).**

Let $\mu$ be a finite Borel measure on $\mathbb{R}^{d}$. A Borel measure $\nu$ is called a Bessel measure for $\mu$, if there exists a positive constant $B$ such that for every $f\in L^{2}(\mu)$,

[TABLE]

Here $B$ is called a (Bessel) bound for $\nu$.

The measure $\nu$ is called a frame measure for $\mu$ if there exist positive constants $A,B$ such that for every $f\in L^{2}(\mu)$,

[TABLE]

In this case, $A$ and $B$ are called (frame) bounds for $\nu$. The measure $\nu$ is called a tight frame measure if $A=B$ and Plancherel measure if $A=B=1$ (see also [8]).

The set of all Bessel measures for $\mu$ with fixed bound $B$ is denoted by $\mathcal{B}_{B}(\mu)$ and the set of all frame measures for $\mu$ with fixed bounds $A,B$ is denoted by $\mathcal{F}_{A,B}(\mu)$.

Remark 2.5**.**

A compactly supported probability measure $\mu$ is an F-spectral measure if and only if there exists a countable set $\Lambda$ in $\mathbb{R}^{d}$ such that $\nu=\sum_{\lambda\in\Lambda}\delta_{\lambda}$ is a frame measure for $\mu$.

Definition 2.6**.**

A finite set of contraction mappings $\{\tau_{i}\}_{i=1}^{n}$ on a complete metric space is called an iterated function system (IFS). Hutchinson [15] proved that, for the metric space $\mathbb{R}^{d}$, there exists a unique compact subset $X$ of $\mathbb{R}^{d}$, which satisfies $X=\bigcup_{i=1}^{n}\tau_{i}(X)$. Moreover, if the IFS is associated with a set of probability weights $\{\rho_{i}\}_{i=1}^{n}$ (i.e., $0<\rho_{i}<1$, $\sum_{i=1}^{n}\rho_{i}=1$), then there exists a unique Borel probability measure $\mu$ supported on $X$ such that $\mu=\sum_{i=1}^{n}\rho_{i}(\mu o\tau^{-1})$. The corresponding $X$ and $\mu$ are called the attractor and the invariant measure of the IFS, respectively. It is well known that the invariant measure is either absolutely continuous or singular continuous with respect to Lebesgue measure. In an affine IFS each $\tau_{i}$ is affine and represented by a matrix. If $R$ is a $d\times d$ expanding integer matrix (i.e., all eigenvalues $\lambda$ satisfy $|\lambda|>1$), and $\mathcal{A}\subset\mathbb{Z}^{d}$, with $\#\mathcal{A}=:N\geq 2$, then the following set (associated with a set of probability weights) is an affine iterated function system.

[TABLE]

Since $R$ is expanding, the maps $\tau_{a}$ are contractions (in an appropriate metric equivalent to the Euclidean one). In some cases, the invariant measure $\mu_{\mathcal{A}}$ is a fractal measure (see [3]). For example singular continuous invariant measures supported on Cantor type sets are fractal measures (see [15, 14]).

Definition 2.7** ([22]).**

(Semi-inner product spaces)

Let $X$ be a vector space over the filed $F$ of complex (real) numbers. If a function $[\cdot,\cdot]:X\times X\rightarrow F$ satisfies the following properties:

$[x+y,z]=[x,z]+[y,z],\;\;\ \text{for}\ x,y,z\in X;$ 2. 2.

$[\lambda x,y]=\lambda[x,y],\;\ \text{for}\ \lambda\in F\ \text{and}\ x,y\in X;$ 3. 3.

$[x,x]>0,\;\ \text{for}\ x\neq 0;$ 4. 4.

$|[x,y]|^{2}\leq[x,x][y,y],$

then $[\cdot,\cdot]$ is called a semi-inner product and the pair$(X,[\cdot,\cdot])$ is called a semi-inner product space. It is easy to observe that $\|x\|=[x,x]^{\frac{1}{2}}$ is a norm on $X$. So every semi-inner product space is a normed linear space. On the other hand, one can generate a semi-inner product in a normed linear space, in infinitely many different ways.

As a matter of fact, semi-inner products provide the possibility of carrying over Hilbert space type arguments to Banach spaces.

Every Banach space has a semi-inner product that is compatible. For example consider the Banach function space $L^{p}(X,\mu),\;p\geq 1$, a compatible semi-inner product on this space is defined by (see [12])

[TABLE]

for every $f,g\in L^{p}(X,\mu)$ with $\|g\|_{L^{p}(\mu)}\neq 0$, and $[f,g]_{L^{p}(\mu)}=0$ for $\|g\|_{L^{p}(\mu)}=0$.

To construct frames in a Hilbert space $H$ the sequence space $l^{2}$ is required. Similarly, to construct frames in a Banach space $X$ one needs a Banach space of scaler valued sequences $X_{d}$ (in fact a BK-space $X_{d}$, see [1] and the references therein). According to Zhang and Zhang [26] frames in Banach spaces can be defined via a compatible semi-inner product in the following way:

Definition 2.8**.**

Let $X$ be a Banach space with a compatible semi-inner product $[\cdot,\cdot]$ and norm $\|\cdot\|_{X}$. Let $X_{d}$ be an associated BK-space with norm $\|\cdot\|_{X_{d}}$. A sequence of elements $\{f_{i}\}_{i\in I}\subseteq X$ is called an $X_{d}$-frame for $X$ if $\{[f,f_{i}]\}_{i\in I}\in X_{d}$ for all $f\in X$, and there exist constants $A,B>0$ such that for every $f\in X$,

[TABLE]

See also [25].

Based on Definition 2.8, we present the next definition. We consider the function space $L^{p}(\mu)$ and the sequence space $l^{q}(I)$ (where $p>1$ and $q$ is the conjugate exponent to $p$) as the Banach space and the BK- space, respectively.

Definition 2.9**.**

Suppose that $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. Let $\mu$ be a finite Borel measure on $\mathbb{R}^{d}$ and let $[\cdot,\cdot]$ be the compatible semi-inner product on $L^{p}(\mu)$ as defined above. We say that a sequence $\{f_{i}\}_{i\in I}$ is a $q$-Bessel for $L^{p}(\mu)$ if there exists a constant $B>0$ such that for every $f\in L^{p}(\mu)$,

[TABLE]

We call B a ($q$-Bessel) bound.

We say the sequence $\{f_{i}\}_{i\in I}$ is a $q$-frame for $L^{p}(\mu)$ if there exist constants $A,B>0$ such that for every $f\in L^{p}(\mu)$,

[TABLE]

We call $A,B$ ($q$-frame) bounds. We call the sequence $\{f_{i}\}_{i\in I}$ a tight $q$-frame if $A=B$ and Parseval $q$-frame if $A=B=1$.

We extend the notions of Bessel and frame measures as follows.

Definition 2.10**.**

Suppose that $1\leq p<\infty,1<q\leq\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. Let $\mu$ be a finite Borel measure on $\mathbb{R}^{d}$, and let $[\cdot,\cdot]$ be the compatible semi-inner product on $L^{p}(\mu)$ as defined above. We say that a Borel measure $\nu$ is a $(p,q)$-Bessel measure for $\mu$, if there exists a constant $B>0$ such that for every $f\in L^{p}(\mu)$,

[TABLE]

and

[TABLE]

We call $B$ a ($(p,q)$-Bessel) bound for $\nu$.

We say the Borel measure $\nu$ is a $(p,q)$-frame measure for $\mu$, if there exist constants $A,B>0$ such that for every $f\in L^{p}(\mu)$,

[TABLE]

and

[TABLE]

We call $A,B$ ($(p,q)$-frame) bounds for $\nu$. If $A=B$, we call the measure $\nu$ a tight $(p,q)$-frame measure and if $A=B=1$, we call it a $(p,q)$-Plancherel measure.

We denote the set of all $(p,q)$-Bessel measures for $\mu$ with fixed bound $B$ by $\mathcal{B}_{B}(\mu)_{p,q}$ and the set of all $(p,q)$-frame measures for $\mu$ with fixed bounds $A,B$ by $\mathcal{F}_{A,B}(\mu)_{p,q}$.

Remark 2.11**.**

Since $[f,e_{t}]_{L^{p}(\mu)}=\int_{R^{d}}f(x)e^{-2\pi it\cdot x}d\mu(x)=\widehat{fd\mu}(t)$ for any $f\in L^{p}(\mu)$ and $t\in\mathbb{R}^{d}$, we can also write $\widehat{fd\mu}(t)$ instead of $[f,e_{t}]_{L^{p}(\mu)}$. If there exists a $(p,q)$-Bessel/frame measure $\nu$ for $\mu$, then the function $T_{\nu}:L^{p}(\mu)\rightarrow L^{q}(\nu)$ defined by $T_{\nu}f=\widehat{fd\mu}$ is linear and bounded. For $p=1$, $q=\infty$, every $\sigma$-finite measure $\nu$ on $\mathbb{R}^{d}$ is a $(1,\infty)$-Bessel measure for $\mu$, since we always have $\|\widehat{fd\mu}\|_{\infty}\leq\|f\|_{L^{1}(\mu)}$. More precisely, $\nu\in\mathcal{B}_{1}(\mu)_{(1,\infty)}$.

Theorem 2.12** ([10]).**

(Riesz-Thorin interpolation theorem) Let $1\leq p_{0},p_{1},q_{0},q_{1}\leq\infty$, where $p_{0}\neq p_{1}$ and $q_{0}\neq q_{1}$, and let $T$ be a linear operator. Suppose that for some measure spaces $(Y,\nu)$, $(X,\mu)$, $T:L^{p_{0}}(X,\mu)\rightarrow L^{q_{0}}(Y,\nu)$ is bounded with norm $C_{0}$, and $T:L^{p_{1}}(X,\mu)\rightarrow L^{q_{1}}(Y,\nu)$ is bounded with norm $C_{1}$. Then for all $\theta\in(0,1)$ and $p,q$ defined by $\dfrac{1}{p}=\dfrac{(1-\theta)}{p_{0}}+\dfrac{\theta}{p_{1}};\ \dfrac{1}{q}=\dfrac{1-\theta}{q_{0}}+\dfrac{\theta}{q_{1}}$, there exists a constant $C$ such that $C\leq C_{0}^{(1-\theta)}C_{1}^{\theta}$ and $T:L^{p}(X,\mu)\rightarrow L^{q}(Y,\nu)$ is bounded with norm $C$.

All measures we consider in this paper, are Borel measures on $\mathbb{R}^{d}$.

3 Existence and Examples

In this section we investigate the existence of $(p,q)$-Bessel and $(p,q)$-frame measures and also the existence of $q$-Bessel and $q$-frame sequences. In addition, we construct examples of measures which admit $(p,q)$-Bessel measures.

Proposition 3.1**.**

Suppose that $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. Let $\mu$ be a finite Borel measure. Then every finite Borel measure $\nu$ is a $(p,q)$-Bessel measure for $\mu$.

Proof.

Take $f\in L^{p}(\mu)$ and $t\in\mathbb{R}^{d}$. Then by applying Holder’s inequality

[TABLE]

Thus,

[TABLE]

Therefore $\nu\in\mathcal{B}_{\mu(\mathbb{R}^{d})\nu(\mathbb{R}^{d})}(\mu)_{(p,q)}$. For $p=1$, $q=\infty$, as we mentioned in Remark 2.11 $\nu\in\mathcal{B}_{1}(\mu)_{(1,\infty)}$. ∎

Proposition 3.2**.**

Suppose that $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. Let $\Lambda\subset\mathbb{R}^{d}$, $\#\Lambda<\infty$ and $\mu$ be a finite Borel measure. Then the finite sequence $\{e_{\lambda}\}_{\lambda\in\Lambda}$ is a $q$-Bessel for $L^{p}(\mu)$.

Proof.

Consider the finite discrete measure $\nu=\sum_{\lambda\in\Lambda}\delta_{\lambda}$. Since

[TABLE]

then the assertion follows from Proposition 3.1. ∎

Remark 3.3**.**

Proposition 3.1 shows that the Bessel bound may change for different measures $\nu$. So if we consider Borel probability measures $\nu$, then we have a fixed Bessel bound $\mu(\mathbb{R}^{d})$ for all $\nu$. Moreover, this Bessel bound does not depend on $p,q$, i.e., for every probability measure $\nu$ we have $\nu\in\mathcal{B}_{\mu(\mathbb{R}^{d})}(\mu)_{(p,q)}$, where $1<p<\infty$ and $q$ is the conjugate exponent to $p$. In addition, we obtain from Proposition 3.1 that for all conjugate exponents $p,q>1$ the set $\mathcal{B}_{\mu(\mathbb{R}^{d})}(\mu)_{p,q}$ is infinite, since there are infinitely many probability measures $\nu$ (such as every measure $\nu=\frac{1}{\lambda(S)}\chi_{S}d\lambda$ where $S\subset\mathbb{R}^{d}$ with the finite Lebesgue measure $\lambda(S)$, every finite discrete measure $\nu=\frac{1}{n}\sum_{a=1}^{n}\delta_{a}$ where $\delta_{a}$ denotes the Dirac measure at the point $a$, every invariant measure obtained from an iterated function system, and others).

Proposition 3.4**.**

Suppose that $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. Let $\nu$ be a finite Borel measure. Then $\nu$ is a $(p,q)$-Bessel measure for every finite Borel measure $\mu$. In addition, $\nu\in\mathcal{B}_{\nu(\mathbb{R}^{d})}(\mu)_{(p,q)}$ for all probability measures $\mu$.

Proof.

See the proof of Proposition 3.1. ∎

Corollary 3.5**.**

Suppose that $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. A finite Borel measure $\nu$ is a $(p,q)$-Bessel measure for a finite Borel measure $\mu$, if and only if $\mu$ is a $(p,q)$-Bessel measure for $\nu$. In particular, every finite Borel measure $\mu$ is a $(p,q)$-Bessel measure to itself.

Proof.

The statements are direct consequences of Propositions 3.1 and 3.4. ∎

Lemma 3.6**.**

Suppose that $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. Let $\mu$ be a finite Borel measure. Then the following assertions hold.

(i) If there exists a countable set $\Lambda$ in $\mathbb{R}^{d}$ such that $\{e_{\lambda}\}_{\lambda\in\Lambda}$ is a $q$-frame for $L^{p}(\mu)$, then $\nu=\sum_{\lambda\in\Lambda}\delta_{\lambda}$ is a $(p,q)$-frame measure for $\mu$.

(ii) If $\nu$ is purely atomic, i.e. $\nu=\sum_{\lambda\in\Lambda}d_{\lambda}\delta_{\lambda}$, and a $(p,q)$-frame measure for the probability measure $\mu$, then $\{\sqrt[q]{d_{\lambda}}\;e_{\lambda}\}_{\lambda\in\Lambda}$ is a q-frame for $L^{p}(\mu)$.

Proof.

(i) Let $\nu=\sum_{\lambda\in\Lambda}\delta_{\lambda}$. Then for all $f\in L^{p}(\mu)$,

[TABLE]

(ii) Since for all $f\in L^{p}(\mu)$,

[TABLE]

∎

Example 3.7**.**

Suppose that $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$. If $f\in L^{p}([0,1]^{d})$, then from the Hausdorff-Young inequality we have $\hat{f}\in l^{q}(\mathbb{Z}^{d})$ and $\|\hat{f}\|_{q}\leq\|f\|_{p}$. Therefore the measure $\nu=\sum_{t\in\mathbb{Z}^{d}}\delta_{t}$ is a $(p,q)$-Bessel measure for $\mu=\chi_{\{[0,1]^{d}\}}dx$. Besides, $\{e_{t}\}_{t\in\mathbb{Z}^{d}}$ is a $q$-Bessel for $L^{p}(\mu)$, since $\sum_{t\in\mathbb{Z}^{d}}|[f,e_{t}]_{L^{p}(\mu)}|^{q}\leq\|f\|_{p}^{q}$, where $1<p\leq 2$ and $q$ is the conjugate exponent to $p$.

Proposition 3.8**.**

Let $1<p\leq 2$ and $q$ is the conjugate exponent to $p$. Let $0<a\leq\phi(x)\leq b<\infty$ on $[0,1]^{d}$ and $\phi_{t}(x):=\phi(x)$ for all $t\in\mathbb{Z}^{d}$. Then $\{\phi_{t}e_{t}\}_{t\in\mathbb{Z}^{d}}$ is a $q$-Bessel for $L^{p}([0,1]^{d})$.

Proof.

Take $f\in L^{p}\left([0,1]^{d}\right)$. We have $\dfrac{1}{\|\phi\|_{p}^{p-2}}\phi^{p-1}f\in L^{p}([0,1]^{d})$, since

[TABLE]

Hence by Example 3.7,

[TABLE]

∎

Corollary 3.9**.**

Suppose that $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. Let $\mu$ be a probability measure. Let $0<a\leq\phi(x)\leq b<\infty$ on $supp\mu$ and $\phi_{i}(x):=\phi(x)$ for all $i\in I$. If $\{f_{i}\}_{i\in I}$ is a $q$-frame for $L^{p}(\mu)$, then $\{\phi_{i}f_{i}\}_{i\in I}$ is also a $q$-frame for $L^{p}(\mu)$ and for every $f\in L^{p}(\mu)$,

[TABLE]

Remark 3.10**.**

Example 3.7 cannot be extended to the case $p>2$, since there exist continuous functions $f$ such that $\sum_{n\in\mathbb{Z}}|[f,e_{n}]_{L^{p}(\mu)}|^{2-\epsilon}=\infty$ for all $\epsilon>0$. An example of such a function is

$f(x)=\sum_{n=2}^{\infty}\dfrac{e^{in\log n}}{n^{1/2}(\log n)^{2}}e^{inx}$ (see [17]). Therefore $\{e_{n}\}_{n\in\mathbb{Z}}$ is not a $q$-Bessel for $L^{p}([0,1])$ and also $\nu=\sum_{n\in\mathbb{Z}}\delta_{n}$ is not a $(p,q)$-Bessel measure for $\mu=\chi_{[0,1]}dx$ where $p>2$.

Proposition 3.11**.**

Suppose that $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. Let $\mu$ be a compactly supported Borel probability measure. Consider two subsets of $\mathbb{R}^{d}$, $\Lambda=\{\lambda_{n}\;:n\in\mathbb{N}\}$ and $\Omega=\{\omega_{n}\;:n\in\mathbb{N}\}$ with the property that there exists a positive constant $C$ such that $|\lambda_{n}-\omega_{n}|\leq C$ for $n\in\mathbb{N}$.

(i) If $\{e_{\lambda_{n}}\}_{n\in\mathbb{N}}$ is a $q$-Bessel for $L^{p}(\mu)$, then $\{e_{\omega_{n}}\}_{n\in\mathbb{N}}$ is a $q$-Bessel too.

(ii) If $\{e_{\lambda_{n}}\}_{n\in\mathbb{N}}$ is a $q$-frame for $L^{p}(\mu)$, then there exists a $\delta>0$ such that if $C\leq\delta$ then $\{e_{\omega_{n}}\}_{n\in\mathbb{N}}$ is a $q$-frame too (see [3]).

Proof.

We need only consider the case, when all $\omega_{n}=\left((\omega_{n})_{1},\ldots,(\omega_{n})_{d}\right)$ differ from $\lambda_{n}=\left((\lambda_{n})_{1},\ldots,(\lambda_{n})_{d}\right)$ just on the first component, then the assertion follows by induction on the number of components.

Let supp$\mu\subseteq[-M,M]^{d}$ for some $M>0$. Let $f\in L^{p}(\mu)$ and $x\in\mathbb{R}^{d}$. The function $\widehat{fd\mu}$ is analytic in each variable $t_{1},\ldots,t_{d}$. Moreover,

[TABLE]

Writing the Taylor expansion at $(\lambda_{n})_{1}$ in the first variable and using Holder’s inequality, for all $n\in\mathbb{N}$,

[TABLE]

Considering the $q$-Bessel $\{e_{\lambda_{n}}\}_{n\in\mathbb{N}}$ with a bound $B$, we obtain

[TABLE]

Then

[TABLE]

Hence by Minkowski’s inequality,

[TABLE]

and this implies that $\{e_{\omega_{n}}\}_{n\in\mathbb{N}}$ is a $q$-Bessel for $L^{p}(\mu)$.

To show that $\{e_{\omega_{n}}\}_{n\in\mathbb{N}}$ is also a $q$-frame for $L^{p}(\mu)$, let $A$ be a lower bound for $\{e_{\lambda_{n}}\}_{n\in\mathbb{N}}$. Take $\delta>0$ small enough such that for $0<C\leq\delta$,

[TABLE]

Then, by Minkowski’s inequality,

[TABLE]

Thus the assertion follows. ∎

Proposition 3.12**.**

Suppose that $1\leq p_{0},p_{1}<\infty$ and $q_{0},q_{1}$ are the conjugate exponents to $p_{0},p_{1}$ respectively. If $\nu$ is a $(p_{0},q_{0})$-Bessel measure and a $(p_{1},q_{1})$-Bessel measure for $\mu$, then $\nu$ is also a $(p,q)$-Bessel measure for $\mu$, where $p_{0}<p<p_{1}$ and $q$ is the conjugate exponent to $p$.

Proof.

If $\nu$ is a $(p_{0},q_{0})$-Bessel measure for $\mu$ with bound $C$ and also a $(p_{1},q_{1})$-Bessel measure with bound $D$, we have

[TABLE]

and

[TABLE]

Now if $\dfrac{1}{p}=\dfrac{(1-\theta)}{p_{0}}+\dfrac{\theta}{p_{1}};\ \dfrac{1}{q}=\dfrac{1-\theta}{q_{0}}+\dfrac{\theta}{q_{1}}$, where $0<\theta<1$ (i.e., $p_{0}<p<p_{1}$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$), then the Riesz-Thorin interpolation theorem yields

[TABLE]

where $B\leq C^{\frac{1}{q_{0}}(1-\theta)}D^{\frac{1}{q_{1}}\theta}$ (Considering the fact that if $p_{0}=1$ and $q_{0}=\infty$, then $C^{\frac{1}{q_{0}}}$ changes to $C$, and if $p_{1}=1$ and $q_{1}=\infty$, then $D^{\frac{1}{q_{1}}}$ changes to $D$). Hence $\nu$ is a $(p,q)$-Bessel measure for $\mu$, where $p_{0}<p<p_{1}$ and $q$ is the conjugate exponent to $p$. ∎

Corollary 3.13**.**

If $\nu$ is a Bessel/frame measure for $\mu$, then $\nu$ is also a $(p,q)$-Bessel measure for $\mu$, where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$.

Proof.

Let $p_{0}=1,q_{0}=\infty,p_{1}=2,q_{1}=2$ in the assumption of Proposition 3.12, then the conclusion follows. ∎

Proposition 3.14**.**

If $\nu\in\mathcal{F}_{A,B}(\mu)$, then for any constant $\alpha>0$, $\nu$ is a frame measure for $\alpha\mu$. More precisely $\nu\in\mathcal{F}_{\alpha A,\alpha B}(\alpha\mu)$.

Proof.

Since $\nu\in\mathcal{F}_{A,B}(\mu)$ for all $f\in L^{2}(\mu)$,

[TABLE]

and we have

[TABLE]

Since $\alpha f\in L^{2}(\mu)$,

[TABLE]

Therefore,

[TABLE]

Hence $\nu\in\mathcal{F}_{\alpha A,\alpha B}(\alpha\mu)$. ∎

Theorem 3.15** ([23]).**

There exist positive constants $c,C$ such that for every set $S\subset\mathbb{R}^{d}$ of finite measure, there is a discrete set $\Lambda\subset\mathbb{R}^{d}$ such that $E(\Lambda)$ is a frame for $L^{2}(S)$ with frame bounds $c|S|$ and $C|S|$, where $|S|$ denotes the measure of $S$.

Theorem 3.16**.**

Let $S$ be a subset (not necessarily bounded) of $\mathbb{R}^{d}$ with finite Lebesgue measure $|S|$. Then the probability measure $\mu=\frac{1}{|S|}\chi_{S}dx$ has an infinite discrete $(p,q)$-Bessel measure $\nu$, where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$.

Proof.

By Theorem 3.15, There are positive constants $c,C$ such that for every set $S\subset\mathbb{R}^{d}$ of finite Lebesgue measure $|S|$, there is a discrete set $\Lambda\subset\mathbb{R}^{d}$ such that $E(\Lambda)$ is a frame for $L^{2}(S)$ with frame bounds $c|S|$ and $C|S|$. Then by considering the upper bound of the frame, we have

[TABLE]

Let $\mu=\frac{1}{|S|}\chi_{S}dx$, and then by Proposition 3.14,

[TABLE]

In addition, $\|\{[f,e_{\lambda}]_{L^{1}(\mu)}\}_{\lambda\in\Lambda}\|_{\infty}\leq\|f\|_{L^{1}(\mu)}$, for every $f$ in $L^{1}(\mu)$. Now if $\dfrac{1}{p}=1-\dfrac{\theta}{2};\ \dfrac{1}{q}=\dfrac{\theta}{2}$, for $0<\theta<1$ (i.e., $1<p<2$ and $q$ is the conjugate exponent to $p$), then the Riesz-Thorin interpolation theorem yields

[TABLE]

where $\mathcal{C}\leq C^{\frac{1}{2}\theta}$. Therefore, $\nu=\sum_{\lambda\in\Lambda}\delta_{\lambda}$ is a $(p,q)$-Bessel measure for $\mu=\frac{1}{|S|}\chi_{S}dx$, and we have $\nu\in\mathcal{B}_{\mathcal{C}^{q}}(\mu)_{(p,q)}$, where $1<p<2$ and $q$ is the conjugate exponent to $p$. Moreover, $\nu\in\mathcal{B}_{C}(\mu)_{(2,2)}$ and $\nu\in\mathcal{B}_{1}(\mu)_{(1,\infty)}$. On the other hand for every $1<p<2$ and $q$ (the conjugate exponent to $p$), $\{e_{\lambda}\}_{\lambda\in\Lambda}$ is a $q$-Bessel for $L^{p}(\mu)$, with bound $\mathcal{C}^{q}$. ∎

If $S\subset\mathbb{R}^{d}$ is a compact set with positive Lebesgue measure, then by Theorem 3.15, we always have the measure $\mu=\frac{1}{|S|}\chi_{S}dx$ is an F-spectral measure, but whether it is a spectral measure, is related to Fuglede’s conjecture [11]. In the following example we consider a spectral measure of this type.

Example 3.17**.**

Let $\mu=\chi_{\left\{[0,1]\cup[2,3]\right\}}dx$. The set of exponential functions $\{e_{\lambda}\;:\lambda\in\Lambda:=\mathbb{Z}\cup\mathbb{Z}+\frac{1}{4}\}$ is an orthogonal basis for $L^{2}(\mu)$ (see [9]). We consider the probability measure $\mu^{\prime}=\dfrac{1}{2}\chi_{\left\{[0,1]\cup[2,3]\right\}}dx$. Then for every $f$ in $L^{2}(\mu^{\prime})$ we have $\sum_{\lambda\in\Lambda}|\left<f,e_{\lambda}\right>_{L^{2}(\mu^{\prime})}|^{2}=\|f\|_{L^{2}(\mu^{\prime})}^{2}$. In addition, for every $f\in L^{1}(\mu^{\prime})$, we have $\|\{[f,e_{\lambda}]_{L^{1}(\mu^{\prime})}\}_{\lambda\in\Lambda}\|_{\infty}\leq\|f\|_{L^{1}(\mu^{\prime})}$. Now by applying the Riesz-Thorin interpolation theorem $\sum_{\lambda\in\Lambda}|[f,e_{\lambda}]_{L^{2}(\mu^{\prime})}|^{q}\leq\|f\|_{L^{p}(\mu^{\prime})}^{q}$, for all $f\in L^{p}(\mu^{\prime})$, where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$. Hence, $\nu=\sum_{\lambda\in\Lambda}\delta_{\lambda}$ is a $(p,q)$-Bessel measure for $\mu^{\prime}$, especially $\nu\in\mathcal{B}_{1}(\mu^{\prime})_{(p,q)}$, where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$. Besides, $\{e_{\lambda}\}_{\lambda\in\Lambda}$ is a $q$-Bessel for $L^{p}(\mu^{\prime})$ with bound $1$, where $1<p\leq 2$ and $q$ is the conjugate exponent to $p$.

Proposition 3.18** ([20]).**

Let $\mu(x)=\phi(x)dx$ be a compactly supported absolutely continuous probability measure. Then $\mu$ is an F-spectral measure if and only if the density function $\phi(x)$ is bounded above and below almost everywhere on the support (see also [8]).

Corollary 3.19**.**

If the density function of a compactly supported absolutely continuous probability measure $\mu$ is essentially bounded above and below on the support, then the following assertions hold.

(i) There exists an infinite $(p,q)$-Bessel measure $\nu=\sum_{\lambda\in\Lambda_{\mu}}\delta_{\lambda}$ for $\mu$, where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$. Moreover, when $\mu$ is a spectral measure, we have $\nu\in\mathcal{B}_{1}(\mu)_{p,q}$, where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$.

(ii) There exists a $q$-Bessel $\{e_{\lambda}\}_{\lambda\in\Lambda_{\mu}}$ for $L^{p}(\mu)$, where $1<p\leq 2$ and $q$ is the conjugate exponent to $p$. In addition, when $\mu$ is a spectral measure, $\{e_{\lambda}\}_{\lambda\in\Lambda_{\mu}}$ is a $q$-Bessel for $L^{p}(\mu)$ with bound $1$, where $1<p\leq 2$ and $q$ is the conjugate exponent to $p$.

Proof.

The conclusion follows from Proposition 3.18 and the Riesz-Thorin interpolation theorem (see the proof of Theorem 3.16 and also, see Example 3.17). ∎

By Proposition 3.4, if $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$, then a fixed finite Borel measure $\nu$ is a $(p,q)$- Bessel measure for every finite measure $\mu$, especially $\nu\in\mathcal{B}_{\nu(\mathbb{R}^{d})}(\mu)_{(p,q)}$ for all probability measures $\mu$, and by Proposition 3.2, if $\nu=\sum_{\lambda\in\Lambda}\delta_{\lambda}$ is a finite discrete $(p,q)$-Bessel measure for $\mu$, then there exists a finite $q$-Bessel for $L^{p}(\mu)$. In the following we give an example of a discrete spectral measure $\mu$ such that it has a finite discrete $(p,q)$-Bessel measure $\nu$ with Bessel bound less than $\nu(\mathbb{R}^{d})$, precisely $\nu\in\mathcal{B}_{1}(\mu)_{(p,q)}$, where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$.

Example 3.20**.**

Consider the atomic measure $\mu:=\frac{1}{2}(\delta_{0}+\delta_{\frac{1}{2}})$, the set $\{e_{l}:\l\in L:=\{0,1\}\}$ is an orthonormal basis for $L^{2}(\mu)$. Hence $\sum_{l\in L}|\left<f,e_{l}\right>_{L^{2}(\mu)}|^{2}=\|f\|_{L^{2}(\mu)}^{2}$ for all $f\in L^{2}(\mu)$. Moreover, for every $f$ in $L^{1}(\mu)$ we have $\|\{[f,e_{l}]_{L^{1}(\mu)}\}_{l\in L}\|_{\infty}\leq\|f\|_{L^{1}(\mu)}$. Now by applying the Riesz-Thorin interpolation theorem $\sum_{l\in L}|[f,e_{l}]_{L^{p}(\mu)}|^{q}\leq\|f\|_{L^{p}(\mu)}^{q}$, for all $f\in L^{p}(\mu)$, where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$. Therefore, $\{e_{l}\}_{l\in L}$ is a finite $q$-Bessel for $L^{p}(\mu)$ with bound 1, and $\nu=\sum_{l\in L}\delta_{l}$ is a finite discrete $(p,q)$-Bessel measure for $\mu$, especially $\nu\in\mathcal{B}_{1}(\mu)_{(p,q)}$, where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$. When $p>2$ and $q$ is the conjugate exponent to $p$, based on Proposition 3.4 $\nu\in\mathcal{B}_{2}(\mu)_{(p,q)}$ and $\{e_{l}\}_{l\in L}$ is a finite $q$-Bessel for $L^{p}(\mu)$ with bound 2.

Proposition 3.21** ([13]).**

Let $\mu=\sum_{c\in C}p_{c}\delta_{c}$ be a discrete probability measure on $\mathbb{R}^{d}$. $\mu$ is an F-spectral measure with an F-spectrum $\Lambda$ if and only if $\#C<\infty$ and $\#\Lambda<\infty$.

Corollary 3.22**.**

Let $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. If $\mu$ is any probability measure, then we have the following two equivalent statements:

(i) A finite discrete measure $\nu=\sum_{\lambda\in\Lambda}\delta_{\lambda}$ is a $(p,q)$-Bessel measure for $\mu$, precisely $\nu\in\mathcal{B}_{\nu(\mathbb{R}^{d})}(\mu)_{(p,q)}$. If $\mu=\sum_{c\in C}p_{c}\delta_{c}$ is an F-spectral measure, then there may exist a better $(p,q)$-Bessel bound for $\nu$, where $1<p\leq 2$ and $q$ is the conjugate exponent to $p$.

(ii) A finite sequence $\{e_{\lambda}\}_{\lambda\in\Lambda}$ is a $q$-Bessel for $L^{p}(\mu)$ with bound $\nu(\mathbb{R}^{d})$. If $\mu=\sum_{c\in C}p_{c}\delta_{c}$ is an F-spectral measure, then $\{e_{\lambda}\}_{\lambda\in\Lambda}$ may admit a better $q$-Bessel bound, where $1<p\leq 2$ and $q$ is the conjugate exponent to $p$.

Proof.

The conclusion follows from Propositions 3.4, 3.21, 3.2, and the Riesz-Thorin interpolation theorem (see also Example 3.20). ∎

Theorem 3.23** ([4]).**

Let $R$ be a $d\times d$ expansive integer matrix, $0\in\mathcal{A}\subset\mathbb{Z}^{d}$. Let $\mu_{\mathcal{A}}$ be an invariant measure associated to the iterated function system

[TABLE]

and the probabilities $(\rho_{a})_{a\in\mathcal{A}}$. Then $\mu$ has an infinite B-spectrum of positive Beurling dimension (Beurling dimension is used as a method of investigating existence of Bessel spectra for singular measures).

Theorem 3.24**.**

Any fractal measure $\mu$ obtained from an affine iterated function system has an infinite discrete $(p,q)$-Bessel measure $\nu$, where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$.

Proof.

Suppose that $R$ is a $d\times d$ expansive integer matrix, $0\in\mathcal{A}\subset\mathbb{Z}^{d}$. If $\mu_{\mathcal{A}}$ is an invariant measure associated to the iterated function system

[TABLE]

and the probabilities $(\rho_{a})_{a\in\mathcal{A}}$, then according to Theorem 3.23 there exist an infinite subset $\Lambda$ of $\mathbb{R}^{d}$ and a constant $B>0$ such that

[TABLE]

We also have $\|\{[f,e_{\lambda}]_{L^{1}(\mu_{\mathcal{A}})}\}_{\lambda\in\Lambda}\|_{\infty}\leq\|f\|_{L^{1}(\mu_{\mathcal{A}})}$, for every $f\in L^{1}(\mu_{\mathcal{A}})$. Now if $\dfrac{1}{p}=1-\dfrac{\theta}{2};\ \dfrac{1}{q}=\dfrac{\theta}{2}$, for $0<\theta<1$ (i.e., $1<p<2$ and $q$ is the conjugate exponent to $p$), then the Riesz-Thorin interpolation theorem yields

[TABLE]

where $B^{\prime}\leq B^{\frac{1}{2}\theta}$. Thus, $\nu=\sum_{\lambda\in\Lambda}\delta_{\lambda}$ is a $(p,q)$-Bessel measure for $\mu_{\mathcal{A}}$, and $\nu\in\mathcal{B}_{B^{\prime q}}(\mu_{\mathcal{A}})_{(p,q)}$, where $1<p<2$ and $q$ is the conjugate exponent to $p$. Moreover, we have $\nu\in\mathcal{B}_{B}(\mu_{\mathcal{A}})_{(2,2)}$ and $\nu\in\mathcal{B}_{1}(\mu_{\mathcal{A}})_{(1,\infty)}$. On the other hand for every $1<p<2$ and $q$ (the conjugate exponent to $p$), $\{e_{\lambda}\}_{\lambda\in\Lambda}$ is a $q$-Bessel for $L^{p}(\mu_{\mathcal{A}})$ with bound $B^{\prime q}$. ∎

If a measure $\mu$ is an F-spectral measure, then it must be of pure type, i.e, $\mu$ is either discrete, singular continuous or absolutely continuous [19, 13]. The case when the measure $\mu$ is singular continuous, is not precisely known. The first known example of a singular continuous spectral measure supported on a non-integer dimension set (a fractal measure), was given by Jorgensen and Pedersen [16]. They showed that the measure $\mu_{4}$ (the Cantor measures supported on Cantor set of $1/4$ contraction), is spectral. A spectrum of $\mu_{4}$ is $\Lambda=\left\{\sum_{m=0}^{k}4^{m}d_{m}\;:d_{m}\in\{0,1\},k\in\mathbb{N}\right\}$. They also showed that $\mu_{2k}$ (the Cantor measures with even contraction ratio) is spectral, but $\mu_{2k+1}$ (the Cantor measures with odd contraction ratio) is not.

Remark 3.25**.**

Since Cantor type measures are fractal measures, by applying Theorem 3.24 one can obtain that every Cantor type measure $\mu$ admits a $(p,q)$-Bessel measure $\nu=\sum_{\lambda\in\Lambda_{\mu}}\delta_{\lambda}$, where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$. Moreover, for every spectral Cantor type measure $\mu_{2k}$, we have $\nu\in\mathcal{B}_{1}(\mu_{2k})_{p,q}$, where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$.

In [21] the author presents a method for constructing many examples of continuous measures $\mu$ (including fractal ones) which have components of different dimensions, but nevertheless they are F-spectral measures. In the following we give some results from [21]. By applying the Riesz-Thorin interpolation theorem, one can obtain infinite discrete $(p,q)$-Bessel measures $\nu=\sum_{\lambda\in\Lambda_{\mu}}\delta_{\lambda}$ (where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$), for such F-spectral measures $\mu$.

Definition 3.26** ([21]).**

Let $\mu$ and $\mu^{\prime}$ be positive and finite measures on $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$, respectively. A mixed type measure $\rho$ is a measure which is constructed on $\mathbb{R}^{n+m}=\mathbb{R}^{n}\times\mathbb{R}^{m}$ and defined by

[TABLE]

where $\delta_{0}$ denotes the Dirac measure at the origin. Equivalently, the measure $\rho$ may be defined by the requirement that

[TABLE]

for every continuous, compactly supported function $f$ on $\mathbb{R}^{n}\times\mathbb{R}^{m}$.

Theorem 3.27** ([21]).**

Let $\mu$ and $\mu^{\prime}$ be continuous F-spectral measures. Then the mixed type measure $\rho=\mu\times\delta_{0}+\delta_{0}\times\mu^{\prime}$ is also an F-spectral measure.

Theorem 3.28** ([21]).**

If $\mu$ is the sum of the $k$-dimensional area measure on $[0,1]^{k}\times\{0\}^{d-k}$, and the

$j$-dimensional area measure on $\{0\}^{d-j}\times[0,1]^{j}$ where $1\leq j,k\leq d-1$, then $\mu$ is an F-spectral measure.

The following theorem provides many examples of single dimensional measures which are F-spectral measures:

Theorem 3.29** ([21]).**

Let $\phi:\mathbb{R}^{k}\rightarrow\mathbb{R}^{d-k}$ be a smooth function $(1\leq k\leq d-1)$. If $\mu$ is the k-dimensional area measure on a compact subset of the graph $\{(x,\phi(x)):x\in\mathbb{R}^{k}\}$ of $\phi$, then $\mu$ is an F-spectral measure.

The next proposition shows that if $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$, then considering any countable subset (finite or infinite) $\Lambda$ of $\mathbb{R}^{d}$, one can obtain tight $(p,q)$-frame measures and $(p,q)$-Plancherel measures $\nu_{\Lambda}$ for $\delta_{0}$. In addition, there exist tight and Parseval $q$-frames for $L^{p}(\delta_{0})$.

Proposition 3.30**.**

Suppose that $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. Then there exists a measure $\mu$ which admits tight $(p,q)$-frame measures and $(p,q)$-Plancherel measures. Moreover, there exist tight and Parseval $q$-frames for $L^{p}(\mu)$.

Proof.

Let $\mu=\delta_{0}$. For a countable subset $\Lambda$ of $\mathbb{R}^{d}$, Let $\nu_{\Lambda}=\sum_{\lambda\in\Lambda}c_{\lambda}\delta_{\lambda}$ where $c_{\lambda}>0$.

If $\sum_{\lambda\in\Lambda}c_{\lambda}=m\neq 1$, then for all $f\in L^{p}(\mu)$,

[TABLE]

If $0<c_{\lambda}<1$ and $\sum_{\lambda\in\Lambda}c_{\lambda}=1$, then for all $f\in L^{p}(\mu)$,

[TABLE]

On the other hand, for all $f\in L^{p}(\mu)$ we have

[TABLE]

Hence If $\sum_{\lambda\in\Lambda}c_{\lambda}=m\neq 1$, then $\{\sqrt[q]{c}_{\lambda}e_{\lambda}\}_{\lambda\in\Lambda}$ is a tight $q$-frame for $L^{p}(\mu)$, and If $0<c_{\lambda}<1$, $\sum_{\lambda\in\Lambda}c_{\lambda}=1$, then $\{\sqrt[q]{c}_{\lambda}e_{\lambda}\}_{\lambda\in\Lambda}$ is a Parseval $q$-frame for $L^{p}(\mu)$. ∎

Proposition 3.31**.**

Let $\mu$ be a finite Borel measure and $B$ be a positive constant. Then there exists a $(p,q)$-Bessel measure $\nu$ for $\mu$ for all $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$, such that $\nu\in\mathcal{B}_{B}(\mu)_{p,q}$. In addition, for every $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$, there exists a $q$-Bessel with bound $B$ for $L^{p}(\mu)$.

Proof.

Let $\nu=\sum_{i\in I}c_{i}\delta_{\lambda_{i}}$ for some $\lambda_{i}\in\mathbb{R}^{d}$ such that $\sum_{i\in I}c_{i}\leq\dfrac{B}{\mu(\mathbb{R}^{d})}$. Let $p>1$ and $f\in L^{p}(\mu)$. If $q$ is the conjugate exponent to $p$, then by applying Holder’s inequality we have

[TABLE]

Hence $\nu\in\mathcal{B}_{B}(\mu)_{p,q}$.

Since

[TABLE]

the second statement follows from (2). ∎

All infinite $(p,q)$-Bessel measures $\nu$ we observed were discrete. Now the question is whether we can find a finite measure $\mu$ which admits a continuous infinite $(p,q)$-Bessel measure $\nu$. The following proposition shows that the answer is affirmative.

Proposition 3.32**.**

If $\nu=\lambda$ is the Lebesgue measure on $\mathbb{R}^{d}$ and $\mu=\lambda|_{[0,1]^{d}}$, then $\lambda$ is a $(p,q)$-Bessel measure for $\mu$ where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$.

Proof.

According to Plancherel’s theorem the following equation is satisfied:

[TABLE]

If $f$ is supported on $[0,1]^{d}$, then

[TABLE]

Moreover, we have $\|\widehat{fd\mu}\|_{\infty}\leq\|f\|_{L^{1}(\mu)}$ for all $f$ in $L^{1}(\mu)$. Now by applying the Riesz-Thorin interpolation theorem

[TABLE]

where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$. Hence $\lambda\in\mathcal{B}_{1}(\mu)_{p,q}$. ∎

Corollary 3.33**.**

The measure $\mu=\lambda|_{[0,1]^{d}}$ has infinite continuous and discrete $(p,q)$-Bessel measures, where $1\leq p\leq 2$ and $q$ is the conjugate exponent to $p$. More precisely, if $\nu_{1}=\sum_{t\in\mathbb{Z}^{d}}\delta_{t}$ and $\nu_{2}=\lambda$, then $\nu_{1},\nu_{2}\in\mathcal{B}_{1}(\mu)_{p,q}$.

Proof.

The conclusion follows from Example 3.7 and Proposition 3.32. ∎

4 Properties and Structural Results

In this section our assertions are based on the results by Dutkay, Han, and Weber from [5]. We generalize the results and give some of the proofs for completeness.

Proposition 4.1**.**

Let $\mu$ be a Borel probability measure. Let $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. If $\nu$ is a $(p,q)$-Bessel measure for $\mu$, then there exists a constant $C$ such that $\nu(K)\leq Cdiam(K)^{d}$ for any compact subset $K$ of $\mathbb{R}^{d}$. Accordingly, $\nu$ is $\sigma$-finite.

Proof.

It is easy to check that $\widehat{d\mu}:\mathbb{R}^{d}\rightarrow\mathbb{C}$ is uniformly continuous and $\widehat{d\mu}(0)=\mu(\mathbb{R}^{d})=1$. So for every $\eta>0$ there exists $\epsilon>0$ such that for $x\in\mathsf{B}(0,\epsilon)$ we have $|\widehat{d\mu}(0)|-|\widehat{d\mu}(x)|\leq|\widehat{d\mu}(0)-\widehat{d\mu}(x)|\leq\eta$, and then $|\widehat{d\mu}(x)|\geq 1-\eta$. If $\delta:=(1-\eta)^{q}$, then $|\widehat{d\mu}(x)|^{q}\geq\delta$ for $x\in\mathsf{B}(0,\epsilon)$. Thus, for any $t\in\mathbb{R}^{d}$,

[TABLE]

Now Let $K\subseteq\mathbb{R}^{d}$ be compact and $r=diam(K)$. Then there exists a point $x=(x_{1},\ldots,x_{d})$ in $\mathbb{R}^{d}$ such that $K\subset\prod_{i=1}^{d}[x_{i}-r,x_{i}+r]$. We may assume that $\epsilon<2r$ and $2r/\epsilon\in\mathbb{N}$. Let $M=2r/\epsilon$. We have $\prod_{i=1}^{d}[x_{i}-r,x_{i}+r]=\bigcup_{\alpha=1}^{M^{d}}C_{\alpha}$ where $C_{\alpha}$s are d-dimensional cubes of side length $\epsilon$. For any $\alpha\in\{1,\ldots,M^{d}\}$, let $t_{\alpha}$ be the center point of $C_{\alpha}$. Then $C_{\alpha}\subset\mathsf{B}(t_{\alpha},\epsilon)$. Now if $C:=(2/\epsilon)^{d}B/\delta$, then

[TABLE]

Hence the assertion follows. ∎

Theorem 4.2**.**

Let $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. Let $B>A>0$. Then the set $\mathcal{F}_{A,B}(\mu)_{p,q}$ is empty for some finite compactly supported Borel measures $\mu$.

Proof.

Let $\mu=\chi_{[0,1]}dx+\delta_{2}$. Suppose $\nu\in\mathcal{F}_{A,B}(\mu)_{p,q}$. Let $f:=\chi_{\{2\}}$. Then $\|f\|_{L^{p}(\mu)}=1$ and $|[f,e_{t}]|=1$ for all $t\in\mathbb{R}$. In addition, the upper bound implies that $\nu(\mathbb{R})\leq B<\infty$. Then from the inner regularity of Borel measures we obtain that for any $\epsilon>0$ there exist a compact set $K\subset\mathbb{R}$ and a positive constant $R$ such that $\nu(\mathbb{R})-\epsilon<K\leq\nu(\mathsf{B}(0,R))$. Therefore $\nu(\mathbb{R}\setminus\mathsf{B}(0,R))<\epsilon$.

Choose some $T$ large, arbitrary and let $g(x):=e^{-2\pi iTx}\chi_{[0,1]}$. Then

[TABLE]

The substitution $z:=-2\pi x$ gives the last equality. Consequently, $|[g,e_{t}]_{L^{p}(\mu)}|^{q}\leq 1$ for all $t\in\mathbb{R}$, and if we take $T\geq 2R$, then for all $t\in(-R,R)$

[TABLE]

Hence from the lower bound we obtain

[TABLE]

Now if $T\rightarrow\infty$ and $\epsilon\rightarrow 0$, then $A=0$. This is a contradiction. ∎

The next proposition shows that if there exists a $(p,q)$-Bessel/frame measure, then many others can be constructed.

Proposition 4.3**.**

Let $\mu$ be a finite Borel measure and $A,B$ be positive constants. Let $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. Then both sets $\mathcal{B}_{B}(\mu)_{p,q}$ and $\mathcal{F}_{A,B}(\mu)_{p,q}$ are convex and closed under convolution with Borel probability measures.

Proof.

Let $\nu_{1},\nu_{2}\in\mathcal{B}_{B}(\mu)_{p,q}$ and $0<\lambda<1$. For all $f\in L^{p}(\mu)$,

[TABLE]

Then $\lambda\nu_{1}+(1-\lambda)\nu_{2}\in\mathcal{B}_{B}(\mu)_{p,q}$. Similarly, if $\nu_{1},\nu_{2}\in\mathcal{F}_{A,B}(\mu)_{p,q}$, then $\lambda\nu_{1}+(1-\lambda)\nu_{2}\in\mathcal{F}_{A,B}(\mu)_{p,q}$.

Let $s\in\mathbb{R}^{d}$. Then for all $f\in L^{p}(\mu)$,

[TABLE]

In addition, Let $\nu\in\mathcal{B}_{B}(\mu)_{p,q}$ and $\rho$ be a Borel probability measure on $\mathbb{R}^{d}$. Then for any $t\in\mathbb{R}^{d}$ and $f\in L^{p}(\mu)$,

[TABLE]

Therefore,

[TABLE]

For $\nu\in\mathcal{F}_{A,B}(\mu)_{p,q}$ one can obtain the lower bound analogously. ∎

Corollary 4.4**.**

Let $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. If there exists a $(p,q)$-Bessel/frame measure for $\mu$, then there exists one which is absolutely continuous with respect to the Lebesgue measure and whose Radon-Nikodym derivative is $C^{\infty}$.

Proof.

Let $\nu$ be a $(p,q)$-Bessel/frame measure for $\mu$. Convoluting $\nu$ with the Lebesgue measure on $[0,1]$ we have

[TABLE]

where $E$ is any Borel subset of $\mathbb{R}$. Thus, we obtained a $(p,q)$-Bessel/frame measure for $\mu$ which is absolutely continuous with respect to the Lebesgue measure.

Now consider the following two propositions from [10].

$(i)$ If $d\mu=fdm$ and $d\nu=gdm$, then $d(\mu\ast\nu)=(f\ast g)dm$.

$(ii)$ If $f\in L^{1}$ (or $f$ is locally integrable on $\mathbb{R}^{d}$), $g\in C^{k}$, and $\partial^{\alpha}$ is bounded for $|\alpha|\leq k$, then $f\ast g\in C^{k}$ and $\partial^{\alpha}(f\ast g)=f\ast(\partial^{\alpha}g)$ for $|\alpha|\leq k$.

Let $g\geq 0$ be a compactly supported $C^{\infty}$-function with $\int g(t)dt=1$. Let $d\mu_{0}=gdm$ and

$d\nu_{0}=\nu\ast\chi_{[0,1]}dm$. Then we have $d(\mu_{0}\ast\nu_{0})=(\nu([\cdot-1,\cdot])\ast g)dm$ and $\nu([\cdot-1,\cdot])\ast g\in C^{\infty}$. ∎

Definition 4.5** ([5]).**

A sequence of Borel probability measures {$\lambda_{n}\}$ is called an approximate identity if

[TABLE]

Lemma 4.6** ([5]).**

Let $\{\lambda_{n}\}$ be an approximate identity. If $f$ is a continuous function on $\mathbb{R}^{d}$, then for any $x\in\mathbb{R}^{d}$, $\int f(x+t)\ d\lambda_{n}(t)\rightarrow f(x)$ as $n\rightarrow\infty$.

By Proposition 4.3, if $\nu$ is a $(p,q)$-Bessel/frame measure for $\mu$, then $\nu\ast\lambda$ is a $(p,q)$-Bessel/frame measure for $\mu$ with the same bound(s), where $\lambda$ is any Borel probability measure. An obvious question is under what conditions the converse is true. The next theorem gives an answer.

Theorem 4.7**.**

Let $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. Let $\{\lambda_{n}\}$ be an approximate identity. Suppose $\nu$ is a $\sigma$-finite Borel measure, and suppose all measures $\nu\ast\lambda_{n}$ are $(p,q)$-Bessel/frame measures for $\mu$ with uniform bounds, independent of $n$. Then $\nu$ is a $(p,q)$-Bessel/frame measure.

Proof.

Take $f\in L^{p}(\mu)$. Since $|[f,e_{.}]_{L^{p}(\mu)}|^{q}$ is continuous on $\mathbb{R}^{d}$, by Lemma 4.6 and Fatou’s lemma we have

[TABLE]

Hence $\nu$ is a $(p,q)$-Bessel measure with the same bound $B$ as $\nu\ast\lambda_{n}$.

Now showing that

[TABLE]

gives the lower bound (see [5]). ∎

We need the following two propositions from [5] to present a general way of constructing $(p,q)$-Bessel/frame measures for a given measure.

Proposition 4.8** ([5]).**

Let $\mu$ and $\mu^{\prime}$ be Borel probability measures. For $f\in L^{1}(\mu)$, the measure $(fd\mu)\ast\mu^{\prime}$ is absolutely continuous w.r.t. $\mu\ast\mu^{\prime}$ and if the Radon-Nikodym derivative is denoted by $Pf$, then

[TABLE]

Proposition 4.9** ([5]).**

Let $\mu$, $\mu^{\prime}$ be two Borel probability measures and $1\leq p\leq\infty$. if $f\in L^{p}(\mu)$, then the function $Pf$ is also in $L^{p}(\mu\ast\mu^{\prime})$ and

[TABLE]

Now we show that if a convolution of two measures admits a $(p,q)$-Bessel/frame measure, then one can obtain a $(p,q)$-Bessel/frame measure for one of the measures in the convolution by using the Fourier transform of the other measure in the convolution.

Proposition 4.10**.**

Let $\mu$, $\mu^{\prime}$ be two Borel probability measures. Let $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. If $\nu$ is a $(p,q)$-Bessel measure for $\mu\ast\mu^{\prime}$, then $|\hat{\mu^{\prime}}|^{q}d\nu$ is a $(p,q)$-Bessel measure for $\mu$ with the same bound.

If in addition $\nu$ is a $(p,q)$-frame measure for $(\mu\ast\mu^{\prime})$ with bounds $A$ and $B$, and for all $f\in L^{p}(\mu)$, $c\|f\|_{L^{p}(\mu)}^{q}\leq\|Pf\|_{L^{p}(\mu\ast\mu^{\prime})}^{q}$, then $|\hat{\mu}^{\prime}|^{q}d\nu$ is a $(p,q)$-frame measure for $\mu$ with bounds $cA$ and $B$.

Proof.

If $\mu,\nu\in M(\mathbb{R}^{d})$, then $\widehat{\mu\ast\nu}=\hat{\mu}.\hat{\nu}$ (see[10]). Take $f\in L^{p}(\mu)$. Then

[TABLE]

Thus, we have

[TABLE]

∎

In the next theorem we have some stability results. In fact, this theorem is a generalization of Proposition 3.11.

Theorem 4.11**.**

Let $\mu$ be a compactly supported Borel probability measure. Let $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. If $\nu$ is a $(p,q)$-Bessel measure for $\mu$, then for any $r>0$ there exists a constant $D>0$ such that

[TABLE]

If $\nu$ is a $(p,q)$-frame measure for $\mu$, then there exist constants $\delta>0$ and $C>0$ such that

[TABLE]

Proof.

The approach is completely similar to the proof of Theorem $2.10$ from [5]. ∎

We show that by using this stability of $(p,q)$-frame measures, one can obtain atomic $(p,q)$-frame measures from a general $(p,q)$-frame measure.

Definition 4.12**.**

Let $Q=[0,1)^{d}$ and $r>0$. If $\nu$ is a Borel measure on $\mathbb{R}^{d}$ and if $(x_{k})_{k\in\mathbb{Z}^{d}}$ is a set of points such that for all $k\in\mathbb{Z}^{d}$ we have $x_{k}\in r(k+Q)$ and $\nu(r(k+Q))<\infty$, then a discretization of the measure $\nu$ is defined by

[TABLE]

Theorem 4.13**.**

Let $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. If a compactly supported Borel probability measure $\mu$ has a $(p,q)$-Bessel/frame measure $\nu$, then it also has an atomic one. More precisely, if $\nu$ is a $(p,q)$-Bessel measure for $\mu$ and if $\nu^{\prime}$ is a discretization of the measure $\nu$, then $\nu^{\prime}$ is a $(p,q)$-Bessel measure for $\mu$.

If $\nu$ is a $(p,q)$-frame measure for $\mu$ and $r>0$ is small enough, then $\nu^{\prime}$ is a $(p,q)$-frame measure for $\mu$.

Proof.

Let $Q=[0,1)^{d}$. Let $(x_{k})_{k\in\mathbb{Z}^{d}}$ be a set of points such that $x_{k}\in r(k+Q)$ for all $k\in\mathbb{Z}^{d}$. For every $x\in r(k+Q)$ define $\epsilon(x):=x_{k}-x$. Thus, $|\epsilon(x)|\leq r\sqrt{d}=:r^{\prime}$ and for any $f\in L^{p}(\mu)$,

[TABLE]

Since we have

[TABLE]

the upper and lower bounds follow from Theorem 4.11. ∎

By Lemma 3.6, if there exists a purely atomic $(p,q)$-frame measure $\nu$ for a probability measure $\mu$, then there exists a $q$-frame for $L^{p}(\mu)$. Now we conclude that if there exists a $(p,q)$-frame measure $\nu$ (not necessarily purely atomic) for a compactly supported probability measure $\mu$, then there exists a $q$-frame for $L^{p}(\mu)$.

Corollary 4.14**.**

Let $\mu$ be a compactly supported Borel probability measure. Let $1<p,q<\infty$ and $\dfrac{1}{p}+\dfrac{1}{q}=1$. If $\nu$ is a $(p,q)$-frame measure for $\mu$ with bounds $A,B$ and $r>0$ is sufficiently small, then there exist positive constants $C,D$ such that $\{c_{k}e_{x_{k}}:k\in\mathbb{Z}^{d}\}$ is a $q$-frame for $L^{p}(\mu)$ with bounds $C,D$, where $x_{k}\in r(k+Q)$ and $c_{k}=\sqrt[q]{\nu(r(k+Q))}$.

Proof.

Let $\nu\in\mathcal{F}_{A,B}(\mu)_{p,q}$. Then by Theorems 4.13 and 4.11, $\nu^{\prime}=\sum_{k\in\mathbb{Z}^{d}}c_{k}^{q}\delta_{x_{k}}$ is a $(p,q)$-frame measure for $\mu$. More precisely, $\nu^{\prime}\in\mathcal{F}_{C,D}(\mu)_{p,q}$. Hence for all $f\in L^{p}(\mu)$,

[TABLE]

∎

Acknowledgements

The authors would like to thank Dr. Nasser Golestani for his valuable guidance and helpful comments.