On the New Description of Exceptional Sets in Asymptotic Estimates for the Entire Functions and the Laplace–Stieltjes Integrals

Abstract

We obtain a new extended description of the exceptional set in the asymptotic Borel-type relation in terms of the maximum of the integrand function for the Laplace–Stieltjes integrals. The obtained description of an exceptional set in the Borel-type relation leaves no room for improvement. In particular, we construct a corresponding measure, a function given by the Laplace–Stieltjes integral with respect to this measure, and a measurable set for which the opposite inequality to the Borel-type relation is fulfilled.

1. Introduction

Analytic functions represented by Laplace–Stieltjes integrals [1,2] have interesting and important properties. If a function $f\left(z\right)$ is represented by the Laplace–Stieltjes integral, then Vogl [3] has proven the uniform convergence on compacts of the Bruwier series constructed by the analytic function f and the uniform approximation of the generating function f by its Bruwier series in cones. Moreover, there are known characterizations [4] of conditionally positive definite functions on Euclidean spaces ${\mathbb{R}}^n$ by certain Laplace–Stielties integrals. The relative growth of the Laplace–Stiltjes-type integrals and the growth properties of the maximum modulus related to the Laplace–Stiltjes-type integrals are investigated in [5,6]. The Laplace–Stieltjes transform is the simple sub-case of the Laplace–Stiltjes-type integral, but there are known inequalities concerning the abscissa of convergence, the abscissa of absolute convergence, and the abscissa of uniform convergence for the transform [7].

There are established lower and upper estimates for the Laplace–Stieltjes integrals [8] with an arbitrary abscissa of convergence. These were used to describe the relationship between the growth of the integral and the maximum of the integrand. Moreover, the lower estimates on a sequence for the maximum of the integrand of Laplace–Stieltjes integrals provide the possibility of deducing an analog of Whittaker’s theorem for a lacunary power series [9].

2. Preliminary Notations and Results

Let ${\mathbb{R}}_{+}=(0,+\infty ).$ For $x,y\in {\mathbb{R}}^p,$ we denote

$$⟨x,y⟩=\sum _{i=1}^p{x}_i{y}_i,\left|x\right|={\left(\sum _{i=1}^p{x}_i^2\right)}^{{\displaystyle \frac{1}{2}}},\parallel x\parallel =\sum _{i=1}^p{x}_i.$$

Let $\mathrm{supp}\nu$ be the support function of a measure $\nu$ in ${\mathbb{R}}^p$, which is a closed set $\mathrm{supp}\nu =E$ such that $\nu ({\mathbb{R}}^p\setminus E)=0$ and $\nu \left(\right\{x\in {\mathbb{R}}^p:|x-x_0|<r\})>0$ for every $x_0\in E$ and $r>0$.

We suppose that $\nu$ is a countably additive nonnegative measure on ${\mathbb{R}}_{+}^p$ with unbounded $\mathrm{supp}\nu$, and $f\left(x\right)$ is an arbitrary nonnegative $\nu$-measurable function on ${\mathbb{R}}_{+}^p.$ By $\nu \left(E\right)$, we denote the $\nu$-measure of a $\nu$-measurable set $E\subset {\mathbb{R}}^p.$

Under ${\mathcal{I}}^p\left(\nu \right)$, we understand the class of the functions $F:{\mathbb{R}}^p\to [0,+\infty )$ admitting such a representation

$$F\left(\sigma \right)={\int }_{{\mathbb{R}}_{+}^p}f\left(x\right)e^{⟨\sigma ,x⟩}\nu \left(dx\right),\sigma \in {\mathbb{R}}^p.$$

(1)

Let us denote by $\mathcal{L}$ the class of nonnegative continuous functions $\psi \left(t\right)$ on $[0,+\infty )$, such that $\psi \left(t\right)\to +\infty$ as $t\to +\infty$, by ${\mathcal{L}}^{+}$ the subclass of non-decreasing functions $\psi \in \mathcal{L}$, and by ${\mathcal{L}}_1$ the subclass of $\mathcal{L}$ of functions $\psi \in \mathcal{L}$, such that ${\int }^{+\infty }{\displaystyle \frac{dt}{\psi \left(t\right)}}<+\infty ;$${\mathcal{L}}_1^{+}={\mathcal{L}}_1\cap {\mathcal{L}}^{+}.$

${\mathcal{L}}_2$ refers to the class-differentiable concave functions $\omega \in {\mathcal{L}}^{+}$ such that

$$\begin{array}{c}{\displaystyle \frac{1}{t}}=O\left({\omega }^{\prime }\left(t\right)\right)(t\to +\infty );\end{array}$$

The problem of finding asymptotic upper estimates for functions F belonging to the class ${\mathcal{I}}^p\left(\nu \right)$ by the asymptotic behavior of the supremum

$$\mu (\sigma ,F)=sup\{f\left(x\right)e^{⟨\sigma ,x⟩}:x\in \mathrm{supp}\nu \}$$

was examined in [10,11] (for $p=1$ in [10]). The results contained only restrictions to the measure $\nu$.

Theorem 6 from [10] states that if for some function $\omega \in {\mathcal{L}}_2$ and ${\nu }_1(a,b]=\nu \left(\{t\in \mathbb{R}:a<t\le b\}\right)$, the condition

$$(\exists {\psi }_1\in {\mathcal{L}}_1^{+},\exists {\psi }_2\in {\mathcal{L}}_1):\underset{t\to +\infty }{\overline{lim}}{\omega }^{\prime }\left({\psi }_1^{-1}\left(t\right)\right)ln{\nu }_1(t-\sqrt{{\psi }_2\left(t\right)};t+\sqrt{{\psi }_2\left(t\right)}]\le d$$

(2)

is satisfied, then for each function $F\in {\mathcal{I}}^1\left(\nu \right)$ it is possible to construct a set $E\subset [0;+\infty )$ of finite Lebesgue measures, such that

$$\begin{array}{c}\omega (lnF(\sigma \left)\right)-\omega (ln\mu (\sigma ,F\left)\right)\le d+o\left(1\right)\end{array}$$

(3)

as $\sigma \to +\infty ,\sigma \notin E$.

A similar multidimensional result for arbitrary $p\ge 2$ was proven in [11,12]. The same problem for the entire Dirichlet series was studied in [13,14]. The growth cones for entire multiple Dirichlet series were first considered by Grechanyuk [15,16]. Since analytic functions are closely related with the harmonic functions, for this class of functions there are known growth estimates in a tangent cone at infinity [17].

Similar cones are base for the tubes B in ${\mathbb{R}}^n+iB\subset {\mathbb{C}}^n$ from [18], for which abstract analytic functions are considered, which are Banach-space-valued or Hilber-space-valued functions. Moreover, the analytic functions can be given as the Fourier–Laplace transform applied to some vector-valued $L^p$ functions.

Throughout this article, each cone $K\subset {\mathbb{R}}^p$ is a real cone with a vertex at the origin $\mathbf{0}=(0,\dots ,0)\in {\mathbb{R}}^p$. For $a<b$ and a measure $\nu$, we define

$${\nu }_0(a,b]=\nu \left(\right\{x\in {\mathbb{R}}^p:a<\parallel x\parallel \le b\left\}\right).$$

For $\alpha >1$ and a measurable set $E\subset {\mathbb{R}}^p$, we denote

$${\tau }_{\alpha }\left(E\right)=\underset{E}{\int }{\displaystyle \frac{d{\sigma }_1\dots d{\sigma }_p}{{\left|\sigma \right|}^{\alpha -1}}}.$$

Theorem 1

([12]). Let K be an arbitrary cone such that its closure without origin is contained in the first p-dimensional octant, i.e., $\overline{K}\setminus \left\{\mathbf{0}\right\}\subset {\mathbb{R}}_{+}^p$. If for a measure ν the condition

$$\begin{array}{c}\underset{0}{\overset{+\infty }{\int }}{\displaystyle \frac{dln{\nu }_0(0,t]}{t}}<+\infty \end{array}$$

(4)

holds, then for each function $F\in {\mathcal{I}}^p\left(\nu \right)$ there exists a set $E\subset {\mathbb{R}}_{+}^p$ satisfies

$${\tau }_p\left(E\right)<+\infty$$

(5)

and such that the relation

$$\begin{array}{c}lnF\left(\sigma \right)\le (1+o(1\left)\right)ln\mu (\sigma ,F)\end{array}$$

(6)

holds as $\left|\sigma \right|\to +\infty$ and $\sigma \in K\setminus E$.

If we substitute $d=0$ and $\omega \left(x\right)=lnx$ in condition (2), then the condition will match with condition (4) (see [10]).

3. Auxiliary Statements: Growth Cones Associated with Functions of Form (1) and Measure

Let us denote by $h_{\nu }\left(\sigma \right)=sup\{⟨x,\sigma ⟩:x\in \mathrm{supp}\nu \},\sigma \in {\mathbb{R}}^p$, that is, the support function of the set $\mathrm{supp}\nu$.

For a function F represented by the Laplace–Stieltjes integral in (1) and for $\sigma \in {\mathbb{R}}^p$, we define the following functions

$$\begin{array}{c}{h}_F\left(\sigma \right)=\underset{t\to +\infty }{\underline{lim}}{\displaystyle \frac{1}{t}}lnF\left(t\sigma \right),h_{\mu }\left(\sigma \right)=\underset{t\to +\infty }{\underline{lim}}{\displaystyle \frac{1}{t}}ln\mu (t\sigma ,F).\end{array}$$

Let us define the following cones in ${\mathbb{R}}^p$ by the equalities

$$\begin{array}{c}\gamma \left(\delta \right)=\{\sigma \in {\mathbb{R}}^p:h_{\delta }\left(\sigma \right)=+\infty \},{\gamma }_0\left(\delta \right)=\{\sigma \in {\mathbb{R}}^p:h_{\delta }\left(\sigma \right)>0\},\end{array}$$

where $\delta$ is either a measure of $\nu$ or one of the functions F and $\mu$.

The following statements are correct. Three of them will be used in the proofs of the main results. Another three lemmas characterize the relationship between the introduced notions of measure cone and growth cone.

Lemma 1

([11], Proposition 1). If F is the function of form (1), where $f\left(x\right)>0$$(x\in {\mathbb{R}}_{+}^p)$, and ν is a countable-additive measure, the support of the measure is unbounded in p-dimensional positive real space (the first octant ${\mathbb{R}}_{+}^p$), then

$$\begin{array}{c}\gamma \left(\nu \right)=\gamma \left(F\right)\subset \gamma \left(\mu \right),{\gamma }_0\left(\nu \right)={\gamma }_0\left(F\right)\subset {\gamma }_0\left(\mu \right).\end{array}$$

Lemma 2

([11], Proposition 2). For every $\sigma \in {\mathbb{R}}^p$

$$h_F\left(\sigma \right)=\underset{t\to +\infty }{lim}{\displaystyle \frac{1}{t}}lnF\left(t\sigma \right)=h_{\nu }\left(\sigma \right).$$

Lemma 3

([11], Proposition 3). If $sup\left\{f\right(x):x\in \mathrm{supp}\nu \}=C<+\infty$, then $\gamma \left(F\right)=\gamma \left(\mu \right)=\gamma \left(\nu \right)$ and ${\gamma }_0\left(F\right)={\gamma }_0\left(\mu \right)={\gamma }_0\left(\nu \right)$.

Lemma 4

([11], Proposition 4). If $sup\left\{f\right(x):x\in \mathrm{supp}\nu \}<+\infty$, then for every $\sigma \in {\mathbb{R}}^p$

$$h_{\mu }\left(\sigma \right)=\underset{t\to +\infty }{lim}{\displaystyle \frac{1}{t}}ln\mu (t\sigma ,F)=h_{\nu }\left(\sigma \right).$$

Lemma 5

([11], Proposition 5). The function $lnF\left(\sigma \right)$ is convex as a function of $\sigma \in {\mathbb{R}}^p$.

Lemma 6

([11], Proposition 5). For every cone K in p-dimensional real space whose vertex is the origin $\mathbf{0}$ and whose closure without the vertex is a subset of the $\gamma \left(F\right)$, i.e., $\overline{K}\setminus \left\{\mathbf{0}\right\}\subset \gamma \left(F\right)$, the asymptotic growth of the Laplace–Stieltjes integral is described as the following

$$\underset{\left|\sigma \right|\to +\infty ,\sigma \in K}{lim}{\displaystyle \frac{lnF\left(\sigma \right)}{\left|\sigma \right|}}=+\infty .$$

Lemma 7.

If F is the function of form (1), where $f\left(x\right)>0$$(x\in {\mathbb{R}}_{+}^p)$, and ν is a countable-additive measure having unbounded support in ${\mathbb{R}}_{+}^p$, then ${\mathbb{R}}_{+}^p\subset \gamma \left(F\right)$.

Proof. 

On the contrary, we assume that there exists $\sigma \in {\mathbb{R}}_{+}^p\setminus \gamma \left(F\right),$$\left|\sigma \right|=1$. Then, for arbitrary $x\in \mathrm{supp}\nu$ we obtain

$$\tilde{\sigma }\parallel x\parallel \le ⟨\sigma ,x⟩\le {\displaystyle \frac{ln\mu \left(t\sigma \right)}{t}}-{\displaystyle \frac{lnf\left(x\right)}{t}}\le C+o\left(1\right)<2C(t\to +\infty ),C<+\infty ,$$

where $\tilde{\sigma }=min\{{\sigma }_j:\sigma =({\sigma }_1,\dots ,{\sigma }_j,\dots {\sigma }_p),1\le j\le p\}>0.$ But, $\parallel x\parallel \ge \left|x\right|$; therefore,

$$\tilde{\sigma }\left|x\right|\le \tilde{\sigma }\parallel x\parallel <2C(t\ge t_0),$$

hence, $\left|x\right|<2C/\tilde{\sigma }$. This is a contradiction with the unboundedness of the support of the measure $\nu$ in ${\mathbb{R}}_{+}^p$. □

4. An Upper Estimate of Logarithhm of the Laplace–Stieltjes Integral in a Cone

We consider the measure $d{\theta }_p=dP\times dt,$ that is the direct product of a probability measure P on the part of the unit sphere $S_1=\{x\in \gamma \left(\mu \right):|x|=1\},$ and the Lebesgue measure on the ray $[0,+\infty ).$ For a ${\theta }_p$-measurable set $E\subset {\mathbb{R}}^p$ we put

$${\theta }_p\left(E\right):=\underset{E}{\int }d{\theta }_p.$$

For the ball $B_p\left(r\right)=\{x\in {\mathbb{R}}^p:\left|x\right|<r\}$ of radius $r>0$, obviously

$${\tau }_p(B_p\left(r\right)\cap {\mathbb{R}}_{+}^p)=r{2}^{-p},{\theta }_p(B_p\left(r\right)\cap {\mathbb{R}}_{+}^p)\ge {\theta }_p(B_p\left(r\right)\cap {\mathbb{R}}_{+}^p)=rP\left(\right\{x>0:\left|x\right|=1\left\}\right),$$

hence, in particular, ${\tau }_p\left({\mathbb{R}}_{+}^p\right)=+\infty ,$${\theta }_p\left({\mathbb{R}}_{+}^p\right)=+\infty$ in the case $P\left(\right\{x>0:\left|x\right|=1\left\}\right)>0.$

For $\sigma \in \gamma \left(\mu \right)$ we define

$${\nu }_{\sigma }\left(t\right)=\nu \left(\{x\in {\mathbb{R}}_{+}^p:0<⟨\sigma ,x⟩\le t\}\right),$$

and also

$$K_{\nu }:=\bigcup _{a>0}{K}_{\nu }\left(a\right),K_{\nu }\left(a\right):=\left\{\sigma \in {\mathbb{R}}_{+}^p:\underset{0}{\overset{+\infty }{\int }}{\displaystyle \frac{dln{\nu }_{\sigma }\left(t\right)}{t}}\le a\right\},a<+\infty .$$

(7)

Denote $E_1=\{x\in {\mathbb{R}}_{+}^p:⟨\sigma ,x⟩\le 0\}$. It is easy to see that from $\sigma \in E_1$ it follows that,

$$F\left(x\right)=\underset{E_1}{\int }f\left(x\right)e^{⟨\sigma ,x⟩}\nu \left(dx\right)+\underset{{\mathbb{R}}_{+}^p\setminus E_1}{\int }f\left(x\right)e^{⟨\sigma ,x⟩}\nu \left(dx\right)\le \underset{{\mathbb{R}}_{+}^p\setminus E_1}{\int }f\left(x\right)e^{⟨\sigma ,x⟩}\nu \left(dx\right)+F\left(0\right).$$

(8)

Lemma 8.

The set $K_{\nu }$ is a real positive cone.

Indeed, if ${\sigma }_0\in K_{\nu }$, then for $\sigma =\alpha {\sigma }_0$, $\alpha >0$, we have

$${\nu }_{\sigma }\left(t\right)=\nu \left(\{x\in {\mathbb{R}}_{+}^p:0<\alpha ⟨{\sigma }_0,x⟩\le t\}\right)=\nu \left(\{x\in {\mathbb{R}}_{+}^p:0<⟨{\sigma }_0,x⟩\le t/\alpha \}\right)={\nu }_{{\sigma }_0}(t/\alpha ),$$

thus

$$\underset{0}{\overset{+\infty }{\int }}{\displaystyle \frac{dln{\nu }_{\sigma }\left(t\right)}{t}}=\underset{0}{\overset{+\infty }{\int }}{\displaystyle \frac{dln{\nu }_{{\sigma }_0}(t/\alpha )}{t}}={\displaystyle \frac{1}{\alpha }}\underset{0}{\overset{+\infty }{\int }}{\displaystyle \frac{dln{\nu }_{{\sigma }_0}\left(u\right)}{u}}<+\infty ,$$

i.e., $\sigma =\alpha {\sigma }_0\in K_{\nu }.$

The following result is the main one in the paper. It claims that the ${\theta }_p$-measure of the intersection of an exceptional set and any real cone is finite. The exceptional set is chosen so that relation (6) holds outside the set.

Theorem 2.

If $F\in {\mathcal{I}}^p\left(\nu \right)$, then there exists a set $E\subset {\mathbb{R}}^p$ satisfying the condition

$$\begin{array}{c}{\theta }_p(E\cap K)<+\infty \end{array}$$

(9)

and such that relation (6) holds as $\left|\sigma \right|\to +\infty ,$$\sigma \in K\setminus E,$ for every $a>0$ and an arbitrary real cone K such that $\overline{K}\setminus \left\{\mathbf{0}\right\}\subset \gamma \left(F\right)\cap K_{\nu }\left(a\right)$.

The next theorem immediately follows from Theorem 2. Condition (7) is not a condition, as denotations were introduced.

Theorem 3.

If $F\in {\mathcal{I}}^p\left(\nu \right)$ and condition (7) holds with ${\nu }_0(a,b]=\nu \left(\right\{x\in {\mathbb{R}}^p:a<\parallel x\parallel \le b\left\}\right)$, then there exists a set E satisfies condition (9) such that relation (6) holds as $\left|\sigma \right|\to +\infty ,\sigma \in K\setminus E,$ for an arbitrary real cone K, such that $\overline{K}\setminus \left\{\mathbf{0}\right\}\subset {\mathbb{R}}_{+}^p$.

Proof of Theorem 3.

We put

$$A=\underset{0}{\overset{+\infty }{\int }}{\displaystyle \frac{dln{\nu }_0(0,t]}{t}}$$

and assume that K is an arbitrary real cone in ${\mathbb{R}}_{+}^p$ with the vertex at the origin, such that $\overline{K}\setminus \left\{\mathbf{0}\right\}\subset {\mathbb{R}}_{+}^p.$. Let

$$y:=inf\{inf\{y_j:y=(y_1,\dots ,y_j,\dots ,y_p),\left|y\right|=1,y\in \overline{K}\}:1\le j\le p\}.$$

Then, $y>0$ and for every $t>0$

$${\nu }_{{\sigma }_0}\left(t\right)=\nu \{x:⟨x,{\sigma }_0⟩\le t\}\le \nu \{x:y\parallel x\parallel \le t\}={\nu }_0(0,t/y].$$

Thus,

$$\overline{K}\setminus \left\{\mathbf{0}\right\}\subset K_{\nu }(A/y).$$

We still need to apply the statement of Theorem 2. □

Proof of Theorem 2.

For ${\sigma }_0\in S_1$, we define

$${\nu }_{{\sigma }_0}(0,t]=\nu \left(\{x\in {\mathbb{R}}_{+}^p:0<⟨{\sigma }_0,x⟩\le t\}\right).$$

Let $F\in {\mathcal{I}}^p\left(\nu \right)$. Consider the function

$$F_0\left(\sigma \right)=\underset{{\mathbb{R}}_{+}^p\setminus E_1}{\int }f\left(x\right)e^{⟨\sigma ,x⟩}\nu \left(dx\right).$$

Without loss of generality, we suppose that $F_0\left(0\right)=1.$

Similar to [11,12], for fixed ${\sigma }_0\in S_1$, we examine the function $g\left(t\right)=ln{F}_0\left(t{\sigma }_0\right),t\in {\mathbb{R}}_{+}.$ By Lemma 5, the function g is convex on a real positive semi-axis. So, the right-hand derivative $g^{\prime }\left(t\right)$ exists at each point $t\in {\mathbb{R}}_{+}$ and is a non-decreasing function. At the same time, the derivative $g^{\prime }\left(t\right)$ exists almost everywhere. In the future, for every instance, instead of $g_{+}^{\prime }\left(t\right)$ we will write $g^{\prime }\left(t\right)$.

Let us consider the probabilistic space $\Omega ={\mathbb{R}}_{+}^p$ with the probabilistic measure

$$\begin{array}{c}{P}_0\left(dx\right)=f\left(x\right)e^{t⟨{\sigma }_0,x⟩}{\displaystyle \frac{\nu \left(dx\right)}{F\left(t{\sigma }_0\right)}}\end{array}$$

and introduce the random variable $\xi =⟨{\sigma }_0,x⟩$ on it. Let us denote by $\mathbf{E}\xi$ the mathematical expectation $\xi$. It is easy to see that $\mathbf{E}\xi =g^{\prime }\left(t\right).$ Similar to [10,11,12] by using Markov’s inequality $P\{\xi >d\}<\mathbf{E}\xi /d$$(d>0)$ with $d=2g^{\prime }\left(t\right)$, we can prove that

$$\begin{array}{c}{F}_0\left(t{\sigma }_0\right)\le 2\mu (t{\sigma }_0,F_0){\nu }_{{\sigma }_0}\left(2g^{\prime }\left(t\right)\right).\end{array}$$

(10)

Similarly to what was carried out in [12], for each fixed ${\sigma }_0,$$|{\sigma }_0|=1,$ we will construct a function $\psi :={\psi }_{{\sigma }_0}\in {\mathcal{L}}_1^{+}$ such that

$$ln{\nu }_{{\sigma }_0}\left(t\right)=o\left({\psi }^{-1}\left(t\right)\right)(t\to +\infty ),\underset{t_0}{\overset{+\infty }{\int }}{\displaystyle \frac{du}{\psi \left(u\right)}}\le 2\sqrt{A}$$

(11)

for some $t_0>0$.

Let $T>t_0>0$, such that ${\nu }_{{\sigma }_0}\left(t_0\right)\ge 1$. We have

$$\begin{array}{c}\underset{t_0}{\overset{T}{\int }}{\displaystyle \frac{dln{\nu }_{{\sigma }_0}\left(t\right)}{t}}={\displaystyle \frac{ln{\nu }_{{\sigma }_0}\left(t\right)}{t}}{|}_{t_0}^T+\underset{t_0}{\overset{T}{\int }}{\displaystyle \frac{ln{\nu }_{{\sigma }_0}\left(t\right)}{t^2}}dt,\end{array}$$

and ${\int }_{t_0}^{+\infty }{\displaystyle \frac{dln{\nu }_{{\sigma }_0}\left(t\right)}{t}}=o\left(1\right)$$(t_0\to +\infty )$. Thus, $\underset{t_0}{\overset{+\infty }{\int }}{\displaystyle \frac{ln{\nu }_{{\sigma }_0}\left(t\right)}{t^2}}dt=o\left(1\right)$$(t_0\to +\infty )$. Hence,

$${\displaystyle \frac{ln{\nu }_{{\sigma }_0}\left(t\right)}{t}}\le 2\underset{t}{\overset{2t}{\int }}{\displaystyle \frac{ln{\nu }_{{\sigma }_0}\left(t\right)}{t^2}}dt=o\left(1\right)(t\to +\infty ).$$

In addition,

$${\displaystyle \frac{ln{\nu }_{{\sigma }_0}\left(t\right)}{t}}\le \underset{t}{\overset{+\infty }{\int }}{\displaystyle \frac{ln{\nu }_{{\sigma }_0}\left(u\right)}{u^2}}du(t>0).$$

For fixed ${\sigma }_0,$$|{\sigma }_0|=1$, we denote

$$\begin{array}{c}l\left(t\right)=\underset{t}{\overset{+\infty }{\int }}{\displaystyle \frac{ln{\nu }_{{\sigma }_0}\left(u\right)}{u^2}}du,L\left(t\right)={\left(l\left(t\right)\right)}^{-{\displaystyle \frac{1}{2}}}(t>0).\end{array}$$

It is easy to see that $l\left(t\right)\searrow 0$$(t\to +\infty )$, thus $L\left(t\right)\nearrow +\infty$$(t\to +\infty ).$ So, we can choose $t_0$ such that for all $t\ge t_0$

$${\displaystyle \frac{ln{\nu }_{{\sigma }_0}\left(t\right)}{t}}\le {\int }_t^{+\infty }{\displaystyle \frac{ln{\nu }_{{\sigma }_0}\left(u\right)}{u^2}}du\le min\{1,A\}.$$

(12)

Now, we choose a positive function $\psi$ increasing to $+\infty$ as $t\to +\infty$, such that the inverse function has the form

$${\psi }^{-1}\left(t\right)=\{\begin{array}{cc}L\left(t\right)ln{\nu }_{{\sigma }_0}\left(t\right),\hfill & \mathrm{if}t\ge t_0,\hfill \\ {\displaystyle \frac{1}{2}}L\left(t_0\right)ln{\nu }_{{\sigma }_0}\left(t_0\right)(1+{\displaystyle \frac{t}{t_0}}),\hfill & \mathrm{if}t\in [0;t_0],\hfill \end{array}$$

where $t_0>0$ is such that $L\left(t_0\right)ln{\nu }_{{\sigma }_0}\left(t_0\right)>0.$ Remark, $l\left(t_0\right)\le 1$. So, $l\left(t_0\right)\le \sqrt{l\left(t_0\right)}$, thus, by inequality (12)

$${\psi }^{-1}\left(t_0\right)={\displaystyle \frac{ln{\nu }_{{\sigma }_0}\left(t_0\right)}{\sqrt{l\left(t_0\right)}}}\le {\displaystyle \frac{ln{\nu }_{{\sigma }_0}\left(t_0\right)}{l\left(t_0\right)}}\le t_0$$

and $t_0\le \psi \left(t_0\right)$. Therefore, by relation (12), we obtain

$$\begin{array}{c}\underset{t_0}{\overset{+\infty }{\int }}{\displaystyle \frac{{\psi }^{-1}\left(t\right)}{t^2}}dt=\underset{t_0}{\overset{+\infty }{\int }}{\displaystyle \frac{L\left(t\right)ln{\nu }_{x_0}\left(t\right)}{t^2}}dt=-\underset{t_0}{\overset{+\infty }{\int }}{\displaystyle \frac{dl\left(t\right)}{\sqrt{l\left(t\right)}}}\\ =2{\left(l\left(t_0\right)\right)}^{1/2}=2(\underset{t_0}{\overset{+\infty }{\int }}{\displaystyle \frac{ln{\nu }_{x_0}\left(t\right)}{t^2}}dt{)}^{1/2}\le 2\sqrt{A}.\end{array}$$

One should observe that $\psi$ is non-decreasing. Then, it is possible to deduce such a lower estimate of the integral ${\displaystyle \underset{T}{\overset{+\infty }{\int }}\frac{{\psi }^{-1}\left(t\right)}{t^2}dt\ge \frac{{\psi }^{-1}\left(T\right)}{T}.}$ Thus, by Cauchy’s criterion, we have $t=o\left(\psi \right(t\left)\right)$$(t\to +\infty ).$ Therefore,

$$\begin{array}{c}\underset{t_0}{\overset{+\infty }{\int }}{\displaystyle \frac{du}{\psi \left(u\right)}}=\underset{\psi \left(t_0\right)}{\overset{+\infty }{\int }}{\displaystyle \frac{d{\psi }^{-1}\left(t\right)}{t}}={\displaystyle \frac{{\psi }^{-1}\left(t\right)}{t}}{|}_{\psi \left(t_0\right)}^{+\infty }+\underset{\psi \left(t_0\right)}{\overset{+\infty }{\int }}{\displaystyle \frac{{\psi }^{-1}\left(t\right)}{t^2}}dt\\ =-{\displaystyle \frac{t_0}{\psi \left(t_0\right)}}+\underset{\psi \left(t_0\right)}{\overset{+\infty }{\int }}{\displaystyle \frac{{\psi }^{-1}\left(t\right)}{t^2}}dt\le 2\sqrt{A}.\end{array}$$

So, we obtain $\psi \in {\mathcal{L}}_1^{+}$ and (11).

Now, note that Lemma 6 implies

$$(t-t_0)g^{\prime }\left(t\right)\ge {\int }_{t_0}^t{g}^{\prime }\left(u\right)du=g\left(t\right)\to +\infty (t\to +\infty )$$

uniformly for ${\sigma }_0\in K$, that is $g^{\prime }\left(t\right)\to +\infty$$(t\to +\infty )$ uniformly for ${\sigma }_0\in K$.

For fixed ${\sigma }_0\in K,$$|{\sigma }_0|=1$, we denote

$$E\left({\sigma }_0\right)=\{x=t{\sigma }_0:t>0,2g^{\prime }\left(t\right)>\psi \left(g\left(t\right)\right)\}$$

and

$$\begin{array}{c}E=\bigcup _{{\sigma }_0\in S_1}E\left({\sigma }_0\right).\end{array}$$

Then, for $x=t{\sigma }_0,x\in K\setminus E$ by definition of the set $E\left({\sigma }_0\right)$

$$\begin{array}{c}ln{\nu }_{{\sigma }_0}\left(2g^{\prime }\left(t\right)\right)\le o\left({\psi }^{-1}\left(2g^{\prime }\left(t\right)\right)\right)\\ =o\left({\psi }^{-1}\left(\psi (ln{F}_0\left(x\right))\right)\right)=o(ln{F}_0\left(x\right))\left(\right|x|\to +\infty ).\end{array}$$

It is easy to see that $\mu (x,F_0)=\mu (x,F)$, and from (8), $F_0\left(x\right)\le F\left(x\right)\le 2F_0\left(x\right)$ for sufficiently large $\left|x\right|$. Therefore, from inequality (10) by definition of the set $E\left({\sigma }_0\right)$, we have

$$\begin{array}{c}lnF\left(x\right)\le ln\left(2F_0\left(x\right)\right)\le ln4+ln\mu (x,F_0)+ln{\nu }_{{\sigma }_0}\left(2g^{\prime }\left(t\right)\right)\\ \le ln\mu (x,F_0)+o(ln{F}_0\left(x\right))=ln\mu (x,F)+o(lnF\left(x\right))\left(\right|x|\to +\infty ).\end{array}$$

It means that the relation (6) is fulfilled as $\left|x\right|\to +\infty$ $(x\in K\setminus E).$

Let us recall that $S_1=\{x\in \gamma \left(\mu \right):|x|=1\}.$ Using Fubini’s theorem and condition (11) for the exceptional set E, we obtain

$${\theta }_p(E\cap K)=\underset{E\cap K}{\int }d{\theta }_p=\underset{S_1\cap K}{\int }\left(\underset{E\left({\sigma }_0\right)}{\int }dt\right)P\left(d{\sigma }_0\right)\le 2\underset{S_1\cap K}{\int }\left(\underset{E\left({\sigma }_0\right)}{\int }{\displaystyle \frac{g^{\prime }\left(t\right)}{\psi \left(g\right(t\left)\right)}}dt\right)P\left(d{\sigma }_0\right)$$

$$\le 2\underset{S_1}{\int }\left(\underset{g\left(0\right)}{\overset{g(+\infty )}{\int }}{\displaystyle \frac{du}{\psi \left(u\right)}}\right)P\left(d{\sigma }_0\right)\le 2\underset{0}{\overset{+\infty }{\int }}{\displaystyle \frac{du}{\psi \left(u\right)}}$$

$$=2{\int }_0^{t_0}{\displaystyle \frac{dt}{\psi \left(t\right)}}+2{\int }_{t_0}^{+\infty }{\displaystyle \frac{dt}{\psi \left(t\right)}}\le 2{\int }_0^{t_0}{\displaystyle \frac{dt}{\psi \left(t\right)}}+4\sqrt{A}<+\infty .$$

Theorem 2 is completely proven. □

5. Existence of the Laplace–Stieltjes Integral with the Greater Growth than the Maximum of the Integrand Function

The necessity of condition (4) in Theorem 2 was proven for a one-dimensional case in [19]. It follows from Theorem 3 ([19]) that if $\nu$ is an arbitrary countably additive measure on ${\mathbb{R}}_{+}$, such that

$$\begin{array}{c}\underset{0}{\overset{+\infty }{\int }}{\displaystyle \frac{dln\nu (0,t]}{t}}=+\infty ,ln\nu (0,t]=O\left(t\right)(t\to +\infty ),\nu (0,t]=\nu \left(\{x\in {\mathbb{R}}_{+}:x\le t\}\right),\end{array}$$

then there exists a constant $d>0$, a point $x_0>0$ and a function $F\in {\mathcal{I}}^1\left(\nu \right)$ represented by the Laplace–Stieltjes integral such that

$$\begin{array}{c}(\forall x\ge x_0):lnF\left(x\right)\ge (1+d)ln\mu (x,F).\end{array}$$

(13)

If the measure $\nu ={\nu }_1\times \dots \times {\nu }_p$$(p\ge 2)$ on ${\mathbb{R}}_{+}^p$ is a direct product of the countably additive measures ${\nu }_j$ on ${\mathbb{R}}_{+}$, then the condition (4) in Theorem 2 is a necessary condition. This fact is contained in the following theorem.

Theorem 4.

Let $\nu ={\nu }_1\times \dots \times {\nu }_p$$(p\ge 2)$, ${\nu }_j$ be the countably additive measures on ${\mathbb{R}}_{+}$, and a measure ${\theta }_p$ be such that ${\theta }_p\left(\right\{x=(x_1,x_2,\dots ,x_p)\in {\mathbb{R}}_{+}^p:\left|x\right|=1,x_j\le x_1(j\in \{2,3,\dots ,p\})\left\}\right)>0$. If $ln{\nu }_0(0,t]=O\left(t\right)(t\to +\infty )$ and condition (4) is not satisfied, then there exist a constant $d>0$, a measurable set E, and a function F from ${\mathcal{I}}^p\left(\nu \right)$, represented by the Laplace–Stieltjes integral with respect to the measure μ, such that for all $x\in E$ inequality (13) holds and ${\theta }_p\left(E\right)=+\infty .$

Proof. 

In Theorem 2, from [12], it was proven if an improper integral in (4) is divergent and the function $ln{\nu }_0(0,t]/t$ is bounded as $t\to +\infty ,$ then there exist a constant $d>0$, a measurable set E, and a function $F\in {\mathcal{I}}^p\left(\nu \right)$, such that for all $x\in E$ inequality (13) holds, where

$$E=\{x\in {\mathbb{R}}_{+}^p:x_1\ge t_0,t_0\le x_j\le x_1,j\in \{2,\dots ,p\}\}.$$

We still now show that ${\theta }_p\left(E\right)=+\infty .$ Let us make sure of this now. Indeed,

$$\begin{array}{c}{\theta }_p\left(E\right)=\underset{E}{\int }d{\theta }_p\ge \underset{S_2}{\int }\underset{t_0{({\prod }_{k=1}^{n-1}sin{\theta }_k)}^{-1}}{\overset{+\infty }{\int }}drP\left(d{x}_0\right)\\ \ge \underset{S_2}{\int }\underset{t_0{({\prod }_{k=1}^{n-1}1/2)}^{-1}}{\overset{+\infty }{\int }}drP\left(d{x}_0\right)=\underset{S_2}{\int }\underset{2^{p-1}{t}_0}{\overset{+\infty }{\int }}drP\left(d{x}_0\right)=+\infty ,\end{array}$$

where

$$\begin{array}{c}{S}_2=\{\left(\prod _{k=1}^{p-1}sin{\theta }_k,cos{\theta }_1\prod _{k=2}^{p-1}sin{\theta }_k,\dots ,cos{\theta }_{p-2}sin{\theta }_{p-1},cos{\theta }_{p-1}\right):\\ {\theta }_1\in [0,\pi /2],{\theta }_m\in [0,\pi /4](m\in \{2,3,\dots ,p-1\})\}.\end{array}$$

Theorem 4 is completely proved. □

6. Non-Improvability of the Description of an Exceptional Set in the Asymptotic Estimates of the Laplace–Stieltjes Integral

Many authors [20] obtained multidimensional analogs of known estimates in Wiman–Valiron’s theory without an explicit description of the exceptional sets. Filevych P. [21] examined such a problem concerning the analytic nature of the exceptional set: what is an impact of the Taylor coefficients of an entire function by the maximum modulus divided by the maximum of the real part for the function? It was discovered that the maximum modulus can have arbitrarily extreme growth with respect to the maximum of the real part or the Nevanlinna characteristic. Many relations in the theory of entire functions [22,23], meromorphic functions [24], and subharmonic functions [25] are valid for points which do not belong to some exceptional set of finite linear or logarithmic measures. Similarly, this fact remains true in the stochastic analysis for random entire functions [26]. Halburd and Korhonen proposed approaches [27] for a slight modification of certain inequalities to remove exceptional sets. For entire functions, there are known exceptional sets of other natures [28]. In particular, there are papers on the exceptional set of algebraic numbers, where a given transcendental entire function takes algebraic values. Such entire functions with rational Taylor coefficients exist for every subset of the field of algebraic numbers closed under complex conjugation [29,30,31].

Under the conditions of Theorem 2, the description (9) of an exceptional set E in the Borel-type relation (6) cannot be improved. To verify this, let us examine multiple Dirichlet series.

We consider the unit Dirac’s measure ${\delta }_{\lambda }\left(G\right)$ of a set G concentrated at the point $\lambda$ and choose the measure $\nu$, such that for each bounded set $G\subset {\mathbb{R}}^p$

$$\nu \left(G\right):=\sum _{\parallel n\parallel =0}^{+\infty }{\delta }_{{\lambda }_n}\left(G\right).$$

(14)

We assume that ${\Lambda }_p={\left({\lambda }_n\right)}_{\parallel n\parallel =1}^{+\infty }$ is a fixed sequence, such that ${\lambda }_n=({\lambda }_{n_1}^{\left(1\right)},\dots ,{\lambda }_{n_p}^{\left(p\right)})$ for $n=(n_1,\dots ,n_p)\in {\mathbb{Z}}_{+}^p$ and $0\le {\lambda }_k^{\left(j\right)}\uparrow +\infty$$(k\to +\infty )$ for all $1\le j\le p.$ Then, $\mathcal{D}\left({\Lambda }_p\right):={\mathcal{I}}^p\left(\nu \right)$ is the class of entire Dirichlet series of the form

$$F\left(z\right)=\sum _{\parallel n\parallel =0}^{+\infty }{a}_n{e}^{⟨z,{\lambda }_n⟩}.$$

(15)

For $F\in \mathcal{D}\left({\Lambda }_p\right)$ and $x\in {\mathbb{R}}_{+}^p$, we denote

$$\begin{array}{c}M(x,F)=sup\left\{\right|F(x+iy)|:y\in {\mathbb{R}}^p\},\mu (x,F)=max\{|a_n|e^{⟨x,{\lambda }_n⟩}:n\in {\mathbb{Z}}_{+}^p\}.\end{array}$$

Theorem 5.

Let $h\in {\mathcal{L}}^{+}$ be an arbitrary function. Then, there exist a countably additive measure ν on the σ-algebra of Borel sets $\mathcal{B}\left({\mathbb{R}}_{+}^p\right),$ satisfying condition (4), a function $F\in {\mathcal{I}}^p\left(\nu \right),$ a constant $d>0$, and a set $E\in \mathcal{B}\left({\mathbb{R}}_{+}^p\right)$ such that:

$$(\forall x\in E):lnF\left(x\right)\ge (1+d)ln\mu (x,F)$$

and

$$\underset{E}{\int }h\left(\right|x\left|\right)d{\theta }_p=+\infty$$

for every measure $d{\theta }_p=dP\times dt$, such that $P\left(S_1^{+}\right)>0$, $S_1^{+}=\{x\in {\mathbb{R}}_{+}^p:\left|x\right|=1\}$.

Proof. 

Let us consider a measure $\nu$ of the form (14). It is obvious that the condition (4) is equivalent to the condition

$$\begin{array}{c}\underset{0}{\overset{+\infty }{\int }}{\displaystyle \frac{dln{n}_0\left(t\right)}{t}}dt<+\infty .\end{array}$$

(16)

It is proved in [32] that for every $h_0\in {\mathcal{L}}^{+}$ there exists a sequence ${\Lambda }_p={\left({\lambda }_n\right)}_{n\in {\mathbb{Z}}_{+}^p}$ satisfying the condition (16), a constant $d>0$, a function $F\in \mathcal{D}\left({\Lambda }_p\right),$ and a set $E\in \mathcal{B}\left({\mathbb{R}}_{+}^p\right)$, such that

$$\begin{array}{c}(\forall x\in E):lnM(x,F)\ge (1+d)ln\mu (x,F),\end{array}$$

(17)

where the set E has the form

$$E=\bigcup _{k=1}^{+\infty }\{(x_1,\dots ,x_p)\in {\mathbb{R}}_{+}^p:r_k\le \parallel x\parallel \le r_k^{\ast }\}(r_k<r_k^{\ast }<r_{k+1},k\in \mathbb{N})$$

and for $E_0={⨆}_{k=1}^{+\infty }(r_k,r_k^{\ast })$

$$\underset{E_0\cap [0;+\infty )}{\int }{h}_0\left(t\right)dt=+\infty .$$

It leaves us to prove that the integral of h over the set E with respect to the measure ${\theta }_p$ is divergent, i.e., $\underset{E}{\int }h\left(\right|x\left|\right)d{\theta }_p=+\infty .$ So, if we recall that $P\left(S_1\right)=1$ for the set $S_1=\{x\in {\mathbb{R}}_{+}^p:\left|x\right|=1\}$, then we obtain

$$\begin{array}{c}\underset{E}{\int }h\left(\right|x\left|\right)d{\theta }_p=\sum _{k=1}^{+\infty }\underset{S_1}{\int }\left(\underset{{\displaystyle \frac{r_k}{\parallel {\sigma }_0\parallel }}}{\overset{{\displaystyle \frac{r_k^{\ast }}{\parallel {\sigma }_0\parallel }}}{\int }}h\left(t\right)dt\right)P\left(d{\sigma }_0\right)\\ =\sum _{k=1}^{+\infty }\underset{S_1}{\int }\left(\underset{r_k}{\overset{r_k^{\ast }}{\int }}h\right({\displaystyle \frac{t}{\parallel {\sigma }_0\parallel }}\left)dt\right){\displaystyle \frac{P\left(d{\sigma }_0\right)}{\parallel {\sigma }_0\parallel }}\\ \ge {\displaystyle \frac{1}{C_p}}\sum _{k=1}^{+\infty }\underset{S_1}{\int }\left(\underset{r_k}{\overset{r_k^{\ast }}{\int }}h\right({\displaystyle \frac{t}{C_p}}\left)dt\right)P\left(d{\sigma }_0\right)\\ ={\displaystyle \frac{P\left(S_1^{+}\right)}{C_p}}\sum _{k=1}^{+\infty }\underset{r_k}{\overset{r_k^{\ast }}{\int }}h\left({\displaystyle \frac{t}{C_p}}\right)dt={\displaystyle \frac{P\left(S_1^{+}\right)}{C_p}}\underset{E_0\cap [0;+\infty )}{\int }{h}_0\left(t\right)dt=+\infty ,\end{array}$$

where $h_0\left(t\right)=h\left({\displaystyle \frac{t}{C_p}}\right),C_p=max\{\parallel x_0\parallel :x_0\in S_1\}\ge 1.$

This completes the proof of Theorem 5. □

Remark 1.

For an arbitrary set $E\subset {\mathbb{R}}_{+}^p$, we write

$$\begin{array}{c}E=\bigcup _{x_0\in S_1}E\left({\sigma }_0\right),\end{array}$$

where $E\left({\sigma }_0\right)=\{x\in E:x=t{\sigma }_0,t>0\}$ for every ${\sigma }_0\in S_1$. It is easy to see that

$$\underset{E}{\int }{\displaystyle \frac{dx}{{\left|x\right|}^{p-1}}}=A\underset{E\left({\sigma }_0\right)\times {\mathbb{R}}_{+}}{\int }P\left(d{\sigma }_0\right)\times dt,$$

where $P\left(d{\sigma }_0\right)$ is a probability measure on $S_1$, and A is the area of $S_1$, i.e., ${\tau }_p\left(E\right)<+\infty ⟹{\theta }_p\left(E\right)<+\infty$ for ${\theta }_p=P\left(d{\sigma }_0\right)\times dt$. On the other hand, for example, in the simplest two-dimensional case $p=2$ we can choose a set $E_0$ such that ${\tau }_2\left(E_0\right)<+\infty$ and construct the set $E=\{x=(x_1,x_2):0<x_1<x_2\}\cup E_0$. Then, for the probabilistic measure P given by $P\left(\right\{x:\left|x\right|=1,x\in E_0\left\}\right)=1$ we obtain ${\theta }_2\left(E\right)<+\infty$. However, it is obvious that ${\tau }_2\left(E\right)=+\infty$.

Then Theorem 3 immediately implies the above-mentioned Theorem 1 ([12]), from Theorem 4, Theorem 2 in [12] follows, and Theorem 5 yields Corollary 1 in [12].

Theorem 2 also improves the relevant statements from [12,13].

7. Discussion

Given the obtained results, we can pose a natural conjecture.

Conjecture 1.

The following statement seems to be correct:

Let a function $\omega \in {\mathcal{L}}_2$. If condition (2) holds with ${\nu }_0(a,b]$ instead of ${\nu }_1(a,b]$, then for each function of form (1) there exists a set E satisfying (9) such that relation (3) holds as $\left|x\right|\to +\infty ,x\in K\setminus E$ for arbitrary cone K, such that $\overline{K}\setminus \left\{\mathbf{0}\right\}\subset {\mathbb{R}}_{+}^p$.

Question 1.

What is an analog of Theorem 2 for asymptotic equality (3)? First of all, it is necessary to find conditions that will ensure the validity of relation (3) in an arbitrary cone $\overline{K}\setminus \left\{\mathbf{0}\right\}\subset \gamma \left(F\right)\cap K_{\nu }.$

The example presented in Remark 1 leads to such a question:

Question 2.

Is there a function $F\in {\mathcal{I}}^p\left(\nu \right)$, given by the Laplace–Stieltjes integral, such that an exceptional set E in the Borel-type relation $lnF\left(\sigma \right)\le (1+o(1\left)\right)ln\mu (\sigma ,F)$ has the properties ${\theta }_p\left(E\right)<+\infty$ and ${\tau }_p\left(E\right)=+\infty ?$.

We do not have an exhaustive answer to the question.

First and foremost, there are two differences that make the results of the paper more (“in-depth”) general:

• In previous results [11,13,32,33], the main estimate of the logarithm of the Laplace–Stieltjes integral is valid in the cones K such that $\overline{K}\setminus \left\{O\right\}\subset {\mathbb{R}}_{+}^p$. In the present paper, the estimate holds for the cones K, satisfying $\overline{K}\setminus \left\{O\right\}\subset \gamma \left(F\right)\subset \gamma \left(\mu \right)$, where $\gamma \left(\mu \right)$ is a growth cone of the function $\mu ,$ i.e., it contains such points $\sigma$ at which the lower limit of $ln\mu (t\sigma ,F)/t$ tends to positive infinity as $t\to +\infty .$ Moreover, there is presented Lemma 7 claiming ${\mathbb{R}}_{+}^p\subset \gamma \left(\mu \right)$. In a general case, one has ${\mathbb{R}}_{+}^p\ne \gamma \left(\mu \right)$ and, for example, if $p=2$, the $\gamma \left(\mu \right)$ can be any angle (cone), such that ${\mathbb{R}}_{+}^p\subset \gamma \left(\mu \right)\subset {\mathbb{R}}^2\setminus \{x=(x_1,x_2):x_1\le 0,x_2\le 0\}.$ Such examples were constructed in papers [15,16].
• Remark 1 claims that the description of the size of the exceptional set obtained by our predecessors [10,11,12] in ${\mathbb{R}}_{+}^p$ follows from the description obtained in this paper, but we obtained the description in a much wider cone $\gamma \left(\mu \right)$.

The main statements of the paper can be treated from the point of view that they provide an almost exhaustive answer to the question: what are the conditions by the measure in the Laplace–Stieltjes integral that provide the validity of the Borel-type relation on the growth cone $\gamma \left(\mu \right)$ outside a set E with ${\theta }_p\left(E\right)<+\infty$? Obviously, if the $\nu \left(dx\right)$ is the Lebesgue measure in ${\mathbb{R}}_{+}^p$, then the restrictions of the obtained theorems are satisfied.