The Wiener–Hopf Equation with Probability Kernel and Submultiplicative Asymptotics of the Inhomogeneous Term

Abstract

We consider the inhomogeneous Wiener–Hopf equation whose kernel is a nonarithmetic probability distribution with positive mean. The inhomogeneous term behaves like a submultiplicative function. We establish asymptotic properties of the solution to which the successive approximations converge. These properties depend on the asymptotics of the submultiplicative function.

1. Introduction

The classical Wiener–Hopf equation has the form

$$z\left(x\right)={\int }_0^{\infty }k(x-y)z\left(y\right)dy+g\left(x\right),x\ge 0,$$

or, equivalently,

$$z\left(x\right)={\int }_{-\infty }^{x}z(x-y)k\left(y\right)dy+g\left(x\right),x\ge 0.$$

We shall consider the inhomogeneous generalized Wiener–Hopf equation

$$z\left(x\right)={\int }_{-\infty }^{x}z(x-y)F\left(dy\right)+g\left(x\right),x\ge 0,$$

(1)

where z is the function sought, F is a given probability distribution on $\mathbb{R}$, and the inhomogeneous term g is a known complex function. A probability distribution G on $\mathbb{R}$ is called nonarithmetic if it is not concentrated on the set of points of the form 0, $\pm \lambda$, $\pm 2\lambda$, … (see Section V.2, Definition 3 of [1]). Let ${\mathbb{R}}_{+}$ be the set of all nonnegative numbers and ${\mathbb{R}}_{-}:=\mathbb{R}\backslash {\mathbb{R}}_{+}$ be the set of all negative numbers. For $c\in \mathbb{C}$, we assume that $c/\infty$ is equal to zero. The relation $a\left(x\right)\sim cb\left(x\right)$ as $x\to \infty$ means that $a\left(x\right)/b\left(x\right)\to c$ as $x\to \infty$; if $c=0$, then $a\left(x\right)=o\left(b\right(x\left)\right)$.

Definition 1.

A positive function $\phi \left(x\right)$, $x\in \mathbb{R}$, is called submultiplicative if it is finite, Borel measurable, and satisfies the conditions: $\phi \left(0\right)=1$, $\phi (x+y)\le \phi \left(x\right)\phi \left(y\right)$, x, $y\in \mathbb{R}$.

The following properties are valid for submultiplicative functions defined on the whole line (Theorem 7.6.2) of [2]:

$$\begin{array}{cc}\hfill -\infty <r_{-}& :=\underset{x\to -\infty }{lim}\frac{log\phi \left(x\right)}{x}=\underset{x<0}{sup}\frac{log\phi \left(x\right)}{x}\hfill \\ \hfill & \le \underset{x>0}{inf}\frac{log\phi \left(x\right)}{x}=\underset{x\to \infty }{lim}\frac{log\phi \left(x\right)}{x}=:r_{+}<\infty .\hfill \end{array}$$

(2)

Here are some examples of submultiplicative function on ${\mathbb{R}}_{+}$: (i) $\phi \left(x\right)={(x+1)}^r$, $r>0$; (ii) $\phi \left(x\right)=exp\left(c{x}^{\beta }\right)$, where $c>0$ and $0<\beta <1$; and (iii) $\phi \left(x\right)=exp\left(\gamma x\right)$, where $\gamma \in \mathbb{R}$. In (i) and (ii), $r_{+}=0$, while in (iii), $r_{+}=\gamma$. The product of a finite number of submultiplicative function is again a submultiplicative function.

In the present paper, we investigate the asymptotic behavior of the solution to Equation (1), where F is a nonarithmetic probability distribution with finite positive mean $\mu :={\int }_{\mathbb{R}}xF\left(dx\right)$ and the function $g\left(x\right)$ is asymptotically equivalent (up to a constant factor) to a nondecreasing submultiplicative function $\phi \left(x\right)$ tending to infinity as $x\to \infty$: $g\left(x\right)\sim c\phi \left(x\right)$ as $x\to \infty$. In the main theorems (Theorems 2 and 3), $\phi \left(x\right)$, $x\in {\mathbb{R}}_{+}$, is a nondecreasing submultiplicative function for which there exists ${lim}_{x\to \infty }\phi (x+y)/\phi \left(x\right)$ for each $y\in \mathbb{R}$. If such a limit exists, then it is equal to $exp\left(r_{+}y\right)$.

Earlier [3], the asymptotic behavior of z was studied in detail under the following assumptions: (i) $\mu \in (0,+\infty ]$ and (ii) g belong to either $g\in L_1(0,\infty )$ or $g\in L_{\infty }(0,\infty )$. Roughly speaking, if $g\in L_1(0,\infty )$, then $z\left(x\right)$ tends to a specific finite limit as $x\to \infty$. Moreover, under appropriate conditions, a submultiplicative rate of convergence was given in the form $o(1/\phi (x\left)\right)$. If $g\in L_{\infty }(0,\infty )$, then $z\left(x\right)=O\left(x\right)$ or even $z\left(x\right)=f(\infty )x/\mu$ as $x\to \infty$, provided $f(\infty ):={lim}_{x\to \infty }f\left(x\right)$ exists.

The existence of the solution to Equation (1) and its explicit form (5) were established in [4] for $g\in L_{\infty }(0,\infty )$ and arbitrary probability distributions F, regardless of whether F is of oscillating or drifting type. If $\mu =0$ and if some other hypotheses are fulfilled, then $z\left(x\right)$ tends to a specific finite limit as $x\to \infty$ (Theorem 4 of [4]).

The stability of an integro-differential equation with a convolution type kernel was studied in [5,6].

2. Preliminaries

Consider the collection $S\left(\phi \right)$ of all complex-valued measures $\varkappa$, such that

$${\parallel \varkappa \parallel }_{\phi }:={\int }_{\mathbb{R}}\phi \left(x\right)\left|\varkappa \right|\left(dx\right)<\infty ;$$

here, $\left|\varkappa \right|$ stands for the total variation of $\varkappa$. The collection $S\left(\phi \right)$ is a Banach algebra with norm ${\parallel ·\parallel }_{\phi }$ by the usual operations of addition and scalar multiplication of measures; the product of two elements $\nu$ and $\varkappa$ of $S\left(\phi \right)$ is defined as their convolution $\nu \ast \varkappa$ (Section 4.16) of [2]. The unit element of $S\left(\phi \right)$ is the measure ${\delta }_0$ of unit mass concentrated at zero. Define the Laplace transform of a measure $\varkappa$ as $\widehat{\varkappa }\left(s\right):={\int }_{\mathbb{R}}exp\left(sx\right)\varkappa \left(dx\right)$. It follows from (2) that the Laplace transform of any $\varkappa \in S\left(\phi \right)$ converges absolutely with respect to $\left|\varkappa \right|$ for all s in the strip $\Pi (r_{-},r_{+}):=\{s\in \mathbb{C}:r_{-}\le \Re s\le r_{+}\}$. Let $\nu$ and $\varkappa$ be two complex-valued measures on the $\sigma$-algebra $\mathcal{B}$ of Borel sets in $\mathbb{R}$. Their convolution is the measure

$$\nu \ast \varkappa \left(A\right):=\underset{\{x+y\in A\}}{\int \int }\nu \left(dx\right)\varkappa \left(dy\right)={\int }_{\mathbb{R}}\nu (A-x)\varkappa \left(dx\right),A\in \mathcal{B},$$

provided the integrals make sense; here, $A-x:=\{y\in \mathbb{R}:x+y\in A\}$. Denote by $F^{n\ast }$ the n-th convolution power of F:

$$F^{0\ast }:={\delta }_0,F^{1\ast }:=F,F^{(n+1)\ast }:=F^{n\ast }\ast F,n\ge 1.$$

Let U be the renewal measure generated by F: $U:={\sum }_{n=0}^{\infty }{F}^{n\ast }$.

Let $X_k$, $k\ge 1$, be independent random variables with the same distribution F not concentrated at zero. These variables generate the random walk $S_0=0$, $S_n=X_1+\dots +X_n$, $n\ge 1$. Put ${\overline{\mathcal{T}}}_{+}:=min\left\{n\ge 1:S_n\ge 0\right\}$. The random variable ${\overline{\mathcal{H}}}_{+}:=S_{{\overline{\mathcal{T}}}_{+}}$ is called the first weak ascending ladder height. Similarly, ${\overline{\mathcal{T}}}_{-}:=min\left\{n\ge 1:S_n<0\right\}$ and ${\overline{\mathcal{H}}}_{-}:=S_{{\overline{\mathcal{T}}}_{-}}$ is the first strong descending ladder height. We have the factorization identity (the symbol $\mathsf{E}$ stands for “expectation”).

$$1-\xi \mathsf{E}\left(e^{s{X}_1}\right)=\left(1-\mathsf{E}\left({\xi }^{{\overline{\mathcal{T}}}_{-}}{e}^{s{\overline{\mathcal{H}}}_{-}}\right)\right)\left(1-\mathsf{E}\left({\xi }^{{\overline{\mathcal{T}}}_{+}}{e}^{s{\overline{\mathcal{H}}}_{+}}\right)\right),\left|\xi \right|\le 1,\Re s=0.$$

(3)

This can easily be deduced from an analogous identity in Section XVIII.3 of [1] for another collection of ladder variables. Denote by $F_{\pm }$ the distributions of the random variables ${\overline{\mathcal{H}}}_{\pm }$, respectively. It follows from the identity (3) that

$${\delta }_0-F=({\delta }_0-F_{-})\ast ({\delta }_0-F_{+}).$$

(4)

Let $U_{\pm }:={\sum }_{k=0}^{\infty }{F}_{\pm }^{k\ast }$ be the renewal measures generated by the distributions $F_{\pm }$, respectively. Denote by ${\mathbf{1}}_{{\mathbb{R}}_{+}}$ the indicator of the subset ${\mathbb{R}}_{+}$ in $\mathbb{R}$: ${\mathbf{1}}_{{\mathbb{R}}_{+}}\left(x\right)=1$ for $x\in {\mathbb{R}}_{+}$ and ${\mathbf{1}}_{{\mathbb{R}}_{+}}\left(x\right)=0$ for $x\in {\mathbb{R}}_{-}$. Extend the function g onto the whole line: $g\left(x\right):=0$, $x<0$. This convention will be valid throughout. Let $\nu$ be a measure defined on $\mathcal{B}$, and $a\left(x\right)$, $x\in \mathbb{R}$, a function. Define the convolution $\nu \ast a\left(x\right)$ as the function ${\int }_{\mathbb{R}}a(x-y)\nu \left(dy\right)$, $x\in \mathbb{R}$. The following theorem has been proven in [4].

Theorem 1.

Let F be a probability distribution and $g\in L_1\left({\mathbb{R}}_{+}\right)$. Then, the function

$$z\left(x\right)=U_{+}\ast \left((U_{-}\ast g){\mathbf{1}}_{{\mathbb{R}}_{+}}\right)\left(x\right),x\in {\mathbb{R}}_{+},$$

(5)

is the solution to Equation (1), which coincides with the solution obtained by successive approximations.

If $\mu$ is finite and positive, then ${\mu }_{+}:={\int }_{\mathbb{R}}x{F}_{+}\left(dx\right)$ is also finite and positive (Section XII.2, Theorem 2 of [1]). We have

$$\mu ={\mu }_{+}(1-F_{-}\left({\mathbb{R}}_{-}\right)),U_{-}({\mathbb{R}}_{-}\cup \left\{0\right\})=\frac{1}{1-F_{-}\left({\mathbb{R}}_{-}\right)}.$$

(6)

In fact, pass in (4) to Laplace transforms and divide both sides by s. We get

$$\frac{1-\widehat{F}\left(s\right)}{s}=\left(1-{\widehat{F}}_{-}\left(s\right)\right)\frac{1-{\widehat{F}}_{+}\left(s\right)}{s},s\ne 0,\Re s=0.$$

Let s tend to zero. Then, the fractions on both sides will tend to $\mu$ and ${\mu }_{+}$, respectively. The second equality in (6) is a consequence of the fact that the distribution $F_{-}$ is defective, i.e., $F_{-}\left({\mathbb{R}}_{-}\right)<1$.

Lemma 1.

Let F be a nonarithmetic probability distribution, such that

$$\mu ={\int }_{\mathbb{R}}xF\left(dx\right)\in (0,\infty )$$

and let $\phi \left(x\right)$, $x\in \mathbb{R}$, be a submultiplicative function with $r_{-}\le 0\le r_{+}$. Assume that

$${\int }_{-\infty }^0\phi \left(x\right)F\left((-\infty ,x]\right)dx<\infty .$$

Suppose additionally that $\widehat{F}\left(r_{-}\right)<1$ if $r_{-}<0$. Then $U_{-}\in S\left(\phi \right)$.

Proof. 

By Theorem 4 in [7] with $n=1$ and Remark 5 therein, we have

$${\int }_{-\infty }^0\phi \left(x\right)F_{-}\left(dx\right)<\infty ,$$

i.e., $F_{-}\in S\left(\phi \right)$. Let us prove that the element $\nu :={\delta }_0-F_{-}$ is invertible in $S\left(\phi \right)$. Let $\nu ={\nu }_{\mathfrak{c}}+{\nu }_{\mathfrak{d}}+{\nu }_{\mathfrak{s}}$ be the decomposition of $\nu$ into absolutely continuous, discrete, and singular components. By Theorem 1 of [8], the element $\nu \in S\left(\phi \right)$ has an inverse if $\widehat{\nu }\left(s\right)\ne 0$ for all $s\in \Pi (r_1,r_2)$, and if

$$\underset{s\in \Pi (r_{-},r_{+})}{inf}\left|\widehat{{\nu }_{\mathfrak{d}}}\left(s\right)\right|>max\left\{\widehat{|{\nu }_{\mathfrak{s}}|}\left(r_{-}\right),\widehat{|{\nu }_{\mathfrak{s}}|}\left(r_{+}\right)\right\}.$$

(7)

Let $F_{-}=F_{-}^{\mathfrak{c}}+F_{-}^{\mathfrak{d}}+F_{-}^{\mathfrak{s}}$ be the decomposition of $F_{-}\in S\left(\phi \right)$ into absolutely continuous, discrete, and singular components. Then, ${\nu }^{\mathfrak{d}}={\delta }_0-F_{-}^{\mathfrak{d}}$ and ${\nu }^{\mathfrak{s}}=-F_{-}^{\mathfrak{s}}$. We have

$$\underset{s\in \Pi (r_{-},r_{+})}{inf}\left|\widehat{{\nu }_{\mathfrak{d}}}\left(s\right)\right|\ge 1-\underset{s\in \Pi (r_{-},r_{+})}{sup}\left|\widehat{F_{-}^{\mathfrak{d}}}\left(s\right)\right|=1-\widehat{F_{-}^{\mathfrak{d}}}\left(r_{-}\right).$$

On the other hand, $max\left\{\widehat{|{\nu }_{\mathfrak{s}}|}\left(r_{-}\right),\widehat{|{\nu }_{\mathfrak{s}}|}\left(r_{+}\right)\right\}=\widehat{F_{-}^{\mathfrak{s}}}\left(r_{-}\right)$. Hence, in order to prove (7), it suffices to show that

$$1-\widehat{F_{-}^{\mathfrak{d}}}\left(r_{-}\right)-\widehat{F_{-}^{\mathfrak{s}}}\left(r_{-}\right)\ge 1-\widehat{F_{-}}\left(r_{-}\right)>0.$$

If $r_{-}=0$, this follows from the fact that the distribution $F_{-}$ is defective. Let $r_{-}<0$. By assumption, $\widehat{F}\left(r_{-}\right)<1$ and, obviously, ${\widehat{F}}_{+}\left(r_{-}\right)<1$. Relation (4) implies

$$1-\widehat{F}\left(s\right)=\left(1-\widehat{F_{-}}\left(s\right)\right)\left(1-\widehat{F_{+}}\left(s\right)\right),s\in \Pi (r_{-},r_{+}),$$

(8)

whence $1-\widehat{F_{-}}\left(r_{-}\right)>0$ and (7) follows. Finally,

$$|\widehat{\nu }\left(s\right)|\ge 1-|\widehat{F_{-}}\left(s\right)|\ge 1-\widehat{F_{-}}\left(\right|s\left|\right)\ge 1-\widehat{F_{-}}\left(r_{-}\right)>0,s\in \Pi (r_{-},r_{+}).$$

Therefore, by Theorem 1 in [8], the measure ${\delta }_0-F_{-}$ is invertible in the Banach algebra $S\left(\phi \right)$ and $U_{-}={({\delta }_0-F_{-})}^{-1}\in S\left(\phi \right)$. The proof of the lemma is complete. □

Lemma 2.

Let $a\left(x\right)$, $x\in {\mathbb{R}}_{+}$, be a monotone nondecreasing positive function. Suppose that ${lim}_{x\to \infty }a(x+y)/a\left(x\right)=1$ for each $y\in \mathbb{R}$. Then,

$$a\left(x\right)=o\left({\int }_0^{x}a\left(y\right)dy\right)asx\to \infty .$$

Proof. 

Let $M>0$ be arbitrary. We have

$${\int }_0^x\frac{a\left(y\right)}{a\left(x\right)}dy\ge {\int }_{x-M}^x\frac{a\left(y\right)}{a\left(x\right)}dy\ge {\int }_{x-M}^x\frac{a(x-M)}{a\left(x\right)}dy=M\frac{a(x-M)}{a\left(x\right)}.$$

It follows that ${lim\; inf}_{x\to \infty }{\int }_0^{x}a\left(y\right)dy/a\left(x\right)=\infty$. The proof of the lemma is complete. □

Lemma 3.

Let G be a nonarithmetic probability distribution on ${\mathbb{R}}_{+}$, such that

$${\mu }_G:={\int }_{\mathbb{R}}xG\left(dx\right)\in (0,\infty )$$

and let $U_G$ be the corresponding renewal measure: $U_G:={\sum }_{n=0}^{\infty }{G}^{n\ast }$. Suppose that $a\left(x\right)$ and $b\left(x\right)$, $x\in {\mathbb{R}}_{+}$, are nonnegative functions such that $a\left(x\right)\sim b\left(x\right)$ as $x\to \infty$. Then,

$$I\left(x\right):=U_G\ast a\left(x\right)\sim U_G\ast b\left(x\right)=:J\left(x\right)asx\to \infty .$$

Proof. 

Given $\epsilon >0$, choose $A>0$, such that

$$(1-\epsilon )b\left(x\right)\le a\left(x\right)\le (1+\epsilon )b\left(x\right),x\ge A.$$

Let

$$I\left(x\right)=\left({\int }_0^{x-A}+{\int }_{x-A}^x\right)a(x-y)U_G\left(dy\right)=:I_1\left(x\right)+I_2\left(x\right).$$

Similarly, let $J\left(x\right)=J_1\left(x\right)+J_2\left(x\right)$. Obviously,

$$1-\epsilon \le \underset{x\to \infty }{lim\; inf}\frac{I_1\left(x\right)}{J_1\left(x\right)}\le \underset{x\to \infty }{lim\; sup}\frac{I_1\left(x\right)}{J_1\left(x\right)}\le 1+\epsilon .$$

Since $\epsilon$ is arbitrary, ${lim}_{x\to \infty }{I}_1\left(x\right)/J_1\left(x\right)=1$, i.e., $I_1\left(x\right)\sim J_1\left(x\right)$ as $x\to \infty$. Moreover, $I_1\left(x\right)\ge a(x-A)U_G\left([0,x-A]\right)\to \infty$ as $x\to \infty$ by the elementary renewal theorem for the measure $U_G$: $U_G\left([0,x]\right)\sim x/{\mu }_G$ as $x\to \infty$ (see Section 1.2 of [9]). According to Blackwell’s theorem (Section XI.1, Theorem 1 of [1]),

$$I_2\left(x\right)\le a\left(A\right)U_G\left((x-A,x]\right)\to a\left(A\right)A/{\mu }_G\mathrm{as}x\to \infty .$$

Hence, $I\left(x\right)\sim I_1\left(x\right)$ as $x\to \infty$. A similar relation also holds for $J\left(x\right)$, which completes the proof of the lemma. □

Lemma 4.

Let $\phi \left(x\right)$, $x\in {\mathbb{R}}_{+}$, be a submultiplicative function, such that there exists $\nu \left(y\right):={lim}_{x\to \infty }\phi (x+y)/\phi \left(x\right)$ for each $y\in \mathbb{R}$. Then $\nu \left(y\right)=exp\left(r_{+}y\right)$, $y\in \mathbb{R}$.

Proof. 

By the Corollary of Theorem 4.17.3 in Section 4.17 of [2], $\nu \left(y\right)=exp\left(\alpha y\right)$ for some $\alpha \in \mathbb{R}$. Given $\epsilon >0$, there exists $n_0=n_0\left(\epsilon \right)$, such that ${\displaystyle log\frac{\phi (n+1)}{\phi \left(n\right)}\le \alpha +\epsilon }$ for $n\ge n_0$.

Hence, $\phi (n_0+m)\le \phi \left(n_0\right)e^{m(\alpha +\epsilon )}$ and

$$r_{+}=\underset{m\to \infty }{lim}\frac{log\phi (n_0+m)}{m}\le \underset{m\to \infty }{lim}\frac{log\phi \left(n_0\right)}{m}+\underset{m\to \infty }{lim}\frac{m(\alpha +\epsilon )}{m}=\alpha +\epsilon .$$

Similarly, $r_{+}\ge \alpha -\epsilon$. Since $\epsilon >0$ is arbitrary, $\alpha =r_{+}$. The proof of the lemma is complete. □

3. Main Results

Theorem 2.

Let F be a nonarithmetic probability distribution, such that

$$\mu ={\int }_{\mathbb{R}}xF\left(dx\right)\in (0,\infty )$$

and let $\phi \left(x\right)$, $x\in {\mathbb{R}}_{+}$, be a nondecreasing continuous submultiplicative function tending to infinity as $x\to \infty$, such that $r_{+}=0$ and there exists ${lim}_{x\to \infty }\phi (x+y)/\phi \left(x\right)$ for each $y\in \mathbb{R}$. Suppose that the inhomogeneous term $g\left(x\right)$, $x\in {\mathbb{R}}_{+}$, is bounded on finite intervals and satisfies the relation $g\left(x\right)\sim c\phi \left(x\right)$ as $x\to \infty$, where $c\in \mathbb{C}$. Assume that

$${\int }_{-\infty }^0\phi \left(\right|x\left|\right)F\left((-\infty ,x]\right)dx<\infty .$$

Then, the function $z\left(x\right)$, $x\in {\mathbb{R}}_{+}$, defined by (5) is a solution to Equation (1) and satisfies the asymptotic relation

$$z\left(x\right)\sim \frac{c}{\mu }{\int }_0^x\phi \left(y\right)dyasx\to \infty .$$

Proof. 

Put $M\left(x\right)={\int }_0^x\phi \left(y\right)dy$. By Lemma 4, ${lim}_{x\to \infty }\phi (x+y)/\phi \left(x\right)=1$ for each $y\in \mathbb{R}$. Extend the function $\phi \left(x\right)$ onto the whole line $\mathbb{R}$ by setting $\phi \left(x\right)=\phi \left(\right|x\left|\right)$ for $x\in {\mathbb{R}}_{-}$. The extended function retains the submultiplicative property and $r_{\pm }=0$. To prove the first statement of the theorem, it suffices to assume $g\ge 0$. Choose $C>0$, such that $g\left(x\right)\le C\phi \left(x\right)$, $x\in {\mathbb{R}}_{+}$. The function $z\left(x\right)$ defined by (5) is finite, since

$$\begin{array}{c}{U}_{-}\ast g\left(x\right)\le C{U}_{-}\ast \phi \left(x\right)=C{\int }_{-\infty }^{0+}\phi (x-y)U_{-}\left(dy\right)\le C\phi \left(x\right){\parallel U_{-}\parallel }_{\phi },\\ z\left(x\right)\le C\parallel U_{-}{\parallel }_{\phi }{\int }_0^x\phi (x-y)U_{+}\left(dy\right)\le C{\parallel U_{-}\parallel }_{\phi }\phi \left(x\right)U_{+}\left([0,x]\right)<\infty \end{array}$$

for all $x\in {\mathbb{R}}_{+}$. Let n be a natural number. Denote by ${\mathbf{1}}_{[0,n]}$ the indicator of $[0,n]$. Consider Equation (1) with the inhomogeneous term $g_n\left(x\right)=g\left(x\right){\mathbf{1}}_{[0,n]}\left(x\right)$. Let $z_n$ be the solution to the equation

$$z_n\left(x\right)={\int }_{-\infty }^x{z}_n(x-y)F\left(dy\right)+g_n\left(x\right),x\in {\mathbb{R}}_{+},$$

(9)

defined by formula (5):

$$z_n\left(x\right)=U_{+}\ast \left((U_{-}\ast g_n){\mathbf{1}}_{{\mathbb{R}}_{+}}\right)\left(x\right),x\in {\mathbb{R}}_{+}.$$

(10)

The integral in (9) can be written as

$${\int }_{\mathbb{R}}{z}_n(x-y){\mathbf{1}}_{[0,x]}\left(y\right)F\left(dy\right)\le z_n\left(x\right)\le z\left(x\right)<\infty .$$

The last two inequalities are consequences of (5). Obviously, $z_n\left(x\right)\uparrow$ as $n\uparrow$. By Section 27, Theorem B of [10], the integral tends to ${\int }_{-\infty }^{x}z(x-y)F\left(dy\right)$ as $n\uparrow \infty$. Letting $n\uparrow \infty$ in (9) and (10), we get that z is a solution to (1). Let us prove the assertion of the theorem for the solution $z_{\phi }$ to (1) for $g=\phi$. Let us show that

$$\frac{U_{-}\ast \phi \left(x\right)}{\phi \left(x\right)}\to U_{-}({\mathbb{R}}_{-}\cup \left\{0\right\})\mathrm{as}x\to \infty .$$

(11)

We have

$$\frac{U_{-}\ast \phi \left(x\right)}{\phi \left(x\right)}={\int }_{-\infty }^0\frac{\phi (x-y)}{\phi \left(x\right)}{U}_{-}\left(dy\right).$$

(12)

By Lemma 4, the integrand tends to 1 as $x\to \infty$ and it is majorized by the $U_{-}$-integrable function $\phi \left(y\right)$, since

$$\frac{\phi (x-y)}{\phi \left(x\right)}\le \phi (-y)=\phi \left(y\right)$$

and $U_{-}\in S\left(\phi \right)$ by Lemma 1. Applying Lebesgue’s bounded convergence theorem (Section 26, Theorem D of [10]), we can pass to the limit under the integral sign in (12), which proves (11). Apply Lemma 3 with the following choice of G, $a\left(x\right)$ and $b\left(x\right)$:

$$G:=F_{+},a\left(x\right):={\mathbf{1}}_{{\mathbb{R}}_{+}}\left(x\right)U_{-}\ast \phi \left(x\right),b\left(x\right):=U_{-}({\mathbb{R}}_{-}\cup \left\{0\right\}){\mathbf{1}}_{{\mathbb{R}}_{+}}\left(x\right)\phi \left(x\right).$$

We get

$$z_{\phi }\left(x\right)={\int }_0^x{U}_{-}\ast \phi (x-y)U_{+}\left(dy\right)\sim U_{-}({\mathbb{R}}_{-}\cup \left\{0\right\}){\int }_0^x\phi (x-y)U_{+}\left(dy\right)\mathrm{as}x\to \infty .$$

Recalling (6), we see that in order to prove the theorem for $z_{\phi }$, it suffices to establish

$$U_{+}\ast \left({\mathbf{1}}_{{\mathbb{R}}_{+}}\phi \right)\left(x\right)={\int }_0^x\phi (x-y)U_{+}\left(dy\right)\sim \frac{1}{{\mu }_{+}}{\int }_0^x\phi \left(y\right)dy=\frac{1}{{\mu }_{+}}M\left(x\right)\mathrm{as}x\to \infty .$$

(13)

Integrating by parts, we get

$$\begin{array}{cc}\hfill {\int }_0^x\phi (x-y)U_{+}\left(dy\right)& =\phi (x-y)U_{+}\left([0,y]\right){|}_{y=0}^x-{\int }_0^x{U}_{+}\left([0,y]\right)d_y\phi (x-y)\hfill \\ \hfill & =U_{+}\left([0,x]\right)-\phi \left(x\right)-{\int }_0^x{U}_{+}\left([0,y]\right)d_y\phi (x-y).\hfill \end{array}$$

(14)

The following three estimates hold:

$$\phi \left(x\right),x,U_{+}\left([0,x]\right)=o\left(M\left(x\right)\right)\mathrm{as}x\to \infty .$$

(15)

The first estimate follows from Lemma 2 with $a\left(x\right)=\phi \left(x\right)$. The second one follows from the assumption $\phi \left(y\right)\to \infty$ as $y\to \infty$. The third estimate follows from the second one and the elementary renewal theorem for the measure $U_{+}$: $U_{+}\left([0,x]\right)\sim x/{\mu }_{+}$ as $x\to \infty$.

Show that

$$\begin{array}{c}-{\int }_0^x{U}_{+}\left([0,y]\right)d_y\phi (x-y)\sim -\frac{1}{{\mu }_{+}}{\int }_0^{x}y{d}_y\phi (x-y)\mathrm{as}x\to \infty ,\end{array}$$

(16)

$$\begin{array}{c}-\frac{1}{{\mu }_{+}}{\int }_0^{x}y{d}_y\phi (x-y)\sim \frac{1}{{\mu }_{+}}M\left(x\right)\mathrm{as}x\to \infty .\end{array}$$

(17)

We prove first (17). This follows from the second estimate in (15) and the equality

$$-{\int }_0^{x}y{d}_y\phi (x-y)=-y\phi (x-y){|}_{y=0}^x+{\int }_0^x\phi (x-y)dy=-x+M\left(x\right).$$

Let $\epsilon >0$ be arbitrary. Use the elementary renewal theorem and choose $y_0=y_0\left(\epsilon \right)$, such that

$$(1-\epsilon )U_{+}\left([0,y]\right)\le \frac{y}{{\mu }_{+}}\le (1+\epsilon )U_{+}\left([0,y]\right),y\ge y_0.$$

Write the left-hand side of (16) in the form

$$-\left({\int }_0^{y_0}+{\int }_{y_0}^x\right)U_{+}\left([0,y]\right)d_y\phi (x-y)=:K_1\left(x\right)+K_2\left(x\right),$$

and let $M_1\left(x\right)+M_2\left(x\right)$ be a similar decomposition for the right-hand side. Obviously,

$$(1-\epsilon )M_2\left(x\right)\le K_2\left(x\right)\le (1+\epsilon )M_2\left(x\right).$$

(18)

Let us prove that, as $x\to \infty$, both sides in (16) are asymptotically equivalent to $K_2\left(x\right)$ and $M_2\left(x\right)$, respectively. We have

$$\begin{array}{cc}\hfill M_2\left(x\right)& =-\frac{1}{{\mu }_{+}}{\int }_{y_0}^{x}y{d}_y\phi (x-y)=-\frac{y}{{\mu }_{+}}\phi (x-y){|}_{y=y_0}^x+\frac{1}{{\mu }_{+}}{\int }_{y_0}^x\phi (x-y)dy\hfill \\ \hfill & =-\frac{x}{{\mu }_{+}}+\frac{y_0}{{\mu }_{+}}\phi (x-y_0)+\frac{1}{{\mu }_{+}}{\int }_0^{x-y_0}\phi \left(y\right)dy.\hfill \end{array}$$

Let us show that

$$M_3\left(x\right):={\int }_0^{x-y_0}\phi \left(y\right)dy\sim M\left(x\right)\mathrm{as}x\to \infty .$$

Using the first estimate in (15), we get

$$\begin{array}{cc}\hfill {\int }_{x-y_0}^x\phi \left(y\right)dy& \le \phi \left(x\right){\int }_{x-y_0}^x\phi (y-x)dy\hfill \\ \hfill & =\phi \left(x\right){\int }_0^{y_0}\phi \left(y\right)dy\le \phi \left(x\right)\phi \left(y_0\right)y_0=o\left(M\left(x\right)\right)\mathrm{as}x\to \infty .\hfill \end{array}$$

Finally,

$$\begin{array}{cc}\hfill \frac{M_3\left(x\right)}{M\left(X\right)}& =\frac{1}{M\left(x\right)}\left({\int }_0^x\phi \left(y\right)dy-{\int }_{x-y_0}^x\phi \left(y\right)dy\right)\hfill \\ \hfill & =1-\frac{1}{M\left(x\right)}{\int }_{x-y_0}^x\phi \left(y\right)dy=1-o\left(1\right)\to 1\mathrm{as}x\to \infty ,\hfill \end{array}$$

which establishes the desired equivalence $M_3\left(x\right)\sim M\left(x\right)$ as $x\to \infty$. Taking into account the estimates in (15), we see that $M_2\left(x\right)\sim M\left(x\right)/{\mu }_{+}$ as $x\to \infty$. Moreover,

$$M_1\left(x\right)=-\frac{-y_0\phi (x-y_0)}{{\mu }_{+}}+\frac{1}{{\mu }_{+}}{\int }_{x-y_0}^x\phi \left(u\right)du.$$

The integral is estimated by $y_0\phi \left(x\right)/{\mu }_{+}$. Thus, $M_1\left(x\right)=o\left(M\left(x\right)\right)$ as $x\to \infty$ (see (15)). Relation (17) is proven. Now, divide all parts of (18) by $M_2\left(x\right)$ and let x tend to infinity. We obtain

$$1-\epsilon \le \underset{x\to \infty }{lim\; inf}\frac{K_2\left(x\right)}{M_2\left(x\right)}\le \underset{x\to \infty }{lim\; sup}\frac{K_2\left(x\right)}{M_2\left(x\right)}\le 1+\epsilon .$$

Hence, $K_2\left(x\right)\sim M_2\left(x\right)\sim M\left(x\right)$ as $x\to \infty$. Relation (16) is proven, since, as $x\to \infty$,

$$\begin{array}{cc}\hfill K_1\left(x\right)& \le -U_{+}\left([0,y_0]\right){\int }_0^{y_0}{d}_y\phi (x-y)\hfill \\ \hfill & =U_{+}\left([0,y_0]\right)[\phi \left(x\right)-\phi (x-y_0)]\le U_{+}\left([0,y_0]\right)\phi \left(x\right)=o\left(M\left(x\right)\right).\hfill \end{array}$$

The equivalence (13) now follows from (14)–(17), which proves the theorem in the particular case $g=\phi$. Let g satisfy the hypotheses of the theorem. If, for some $C>0$, $\left|g\right(x\left)\right|\le C\phi \left(x\right)$, $x\in {\mathbb{R}}_{+}$, then

$$\underset{x\to \infty }{lim\; sup}\left|z\left(x\right)\right|/{\int }_0^x\phi \left(y\right)dy\le \frac{C}{\mu }.$$

It follows that if $c=0$, then $z\left(x\right)=o\left(z_{\phi }\left(x\right)\right)$ as $x\to \infty$. To see this, choose a small $\epsilon >0$ and a natural number n, such that $\left|g\right(x\left)\right|\le \epsilon \phi \left(x\right)$, $x\ge n$. Write

$$g={\mathbf{1}}_{[0,n]}g+(g-{\mathbf{1}}_{[0,n]}g)=:g_1+g_2.$$

Let $z_1$ and $z_2$ be the solutions to (1) corresponding to $g_1$ and $g_2$, respectively. Then, $z=z_1+z_2$ and $|z_2\left(x\right)|\le \epsilon z_{\phi }\left(x\right)$, $x\in {\mathbb{R}}_{+}$. By Theorem 6.2 in [3], $z_1\left(x\right)=o\left(x\right)$ as $x\to \infty$. Since $\phi \left(x\right)\ge 1$, $x\in {\mathbb{R}}_{+}$, it follows that $z_1\left(x\right)=o\left({\int }_0^x\phi \left(y\right)dy\right)$ as $x\to \infty$. Therefore,

$$\underset{x\to \infty }{lim\; sup}\left|z\left(x\right)\right|/{\int }_0^x\phi \left(y\right)dy\le \frac{\epsilon }{\mu }.$$

Since $\epsilon >0$ is arbitrary, the assertion of the theorem is true for $c=0$. Let $c\ne 0$. Write g in the form $g=c\phi +g_1$. Then, $g_1\left(x\right)=o\left(\phi \left(x\right)\right)$ as $x\to \infty$, and we have $z=c{z}_{\phi }+z_1$, where $z_1$ is the solution to Equation (1) with the inhomogeneous term $g_1$. The proof of the theorem is complete. □

Theorem 3.

Let F be a nonarithmetic probability distribution, such that

$$\mu ={\int }_{\mathbb{R}}xF\left(dx\right)\in (0,\infty ),$$

and let $\phi \left(x\right)$, $x\in {\mathbb{R}}_{+}$, be a nondecreasing submultiplicative function, such that $r_{+}>0$, and there exists ${lim}_{x\to \infty }\phi (x+y)/\phi \left(x\right)$ for each $y\in \mathbb{R}$. Suppose that the inhomogeneous term $g\left(x\right)$, $x\in {\mathbb{R}}_{+}$, is bounded on finite intervals and satisfies the relation $g\left(x\right)\sim c\phi \left(x\right)$ as $x\to \infty$, where $c\in \mathbb{C}$. Assume that

$${\int }_{-\infty }^0\phi \left(\right|x\left|\right)F\left((-\infty ,x]\right)dx<\infty$$

and $\widehat{F}(-r_{+})<1$. Then, the function $z\left(x\right)$, $x\in {\mathbb{R}}_{+}$, defined by (5) is a solution to Equation (1) and satisfies the asymptotic relation

$$z\left(x\right)\sim \frac{c}{1-\widehat{F}(-r_{+})}\phi \left(x\right)asx\to \infty .$$

Proof. 

As in the proof of the preceding theorem, we verify that $z\left(x\right)$ is a solution to (1). First, let us prove the assertion of the theorem for the solution $z_{\phi }$ to (1) corresponding to $g=\phi$, i.e., let us prove that, as $x\to \infty$,

$$\frac{z_{\phi }\left(x\right)}{\phi \left(x\right)}={\int }_0^x\frac{U_{-}\ast \phi (x-y)}{\phi \left(x\right)}{U}_{+}\left(dy\right)\to {\widehat{U}}_{-}(-r_{+}){\widehat{U}}_{+}(-r_{+})=\frac{1}{1-\widehat{F}(-r_{+})}.$$

(19)

Write the integrand in the form

$$I(x,y):={\mathbf{1}}_{[0,x]}\left(y\right)\frac{U_{-}\ast \phi (x-y)}{\phi (x-y)}\frac{\phi (x-y)}{\phi \left(x\right)},y\in {\mathbb{R}}_{+}.$$

Notice that

$$\frac{U_{-}\ast \phi \left(x\right)}{\phi \left(x\right)}={\int }_{-\infty }^0\frac{\phi (x-y)}{\phi \left(x\right)}{U}_{-}\left(dy\right)\to {\widehat{U}}_{-}(-r_{+})\mathrm{as}x\to \infty .$$

(20)

In fact, $\phi (x-y)/\phi \left(x\right)\to e^{-r_{+}y}$ as $x\to \infty$ by Lemma 4 and, according to Lemma 1, this ratio is majorized by the $U_{-}$-integrable function $\phi \left(y\right)$, $y\in {\mathbb{R}}_{-}$:

$$\frac{U_{-}\ast \phi \left(x\right)}{\phi \left(x\right)}={\int }_{-\infty }^0\frac{\phi (x-y)}{\phi \left(x\right)}{U}_{-}\left(dy\right)\le {\int }_{-\infty }^0\phi \left(\right|y\left|\right)U_{-}\left(dy\right)={\parallel U_{-}\parallel }_{\phi }<\infty .$$

Relation (20) now follows from Lebesgue’s bounded convergence theorem. Our further actions are as follows. We will pick out a majorant for the function $I(x,y)$, $y\in {\mathbb{R}}_{+}$, in the form $M{e}^{\beta y}$ with $\beta \in (-r_{+},0)$. Then, by Lebesgue’s theorem, we pass to the limit under the integral sign in the left-side integral in (19) as $x\to \infty$, and thus prove relation (19). Put $f\left(x\right)=log\phi \left(x\right)-r_{+}x$. By hypothesis, we have

$$f(x-y)-f\left(x\right)=log\phi (x-y)-log\phi \left(x\right)+r_{+}x\to 0\mathrm{as}x\to \infty$$

(21)

for each $y\in \mathbb{R}$. According to Lemma 1.1 in [11], relation (21) is fulfilled uniformly in $y\in [0,1]$. Hence,

$$\frac{\phi (x-y)exp\left(r_{+}y\right)}{\phi \left(x\right)}\to 1\mathrm{as}x\to \infty$$

uniformly in $y\in [0,1]$. Choose a small $\epsilon >0$ such that $\beta :=log(1+\epsilon )-r_{+}<0$. Let $N=N\left(\epsilon \right)>0$ be an integer such that

$$\frac{\phi (x-y)exp\left(r_{+}y\right)}{\phi \left(x\right)}\le 1+\epsilon ,x\ge N,y\in [0,1].$$

Denote by $\left[x\right]$ the integral part of a real number x; i.e., $\left[x\right]$ is the maximal integer not exceeding x: $x=\left[x\right]+\vartheta$, $\vartheta \in [0,1)$. For $y\in [l,l+1]$, $l=0,\dots ,\left[x\right]-N-1$, we have

$$\begin{array}{c}\frac{\phi (x-y)}{\phi \left(x\right)}=\frac{\phi (x-l-(y-l\left)\right)}{\phi (x-l)}\frac{\phi (x-l)}{\phi \left(x\right)},\\ \frac{\phi (x-l-(y-l\left)\right)}{\phi (x-l)}\le (1+\epsilon )exp(-r_{+}(y-l)),\\ \frac{\phi (x-l)}{\phi \left(x\right)}=\frac{\phi (x-l)}{\phi (x-l+1)}\frac{\phi (x-l+1)}{\phi (x-l+2)}\dots \frac{\phi (x-1)}{\phi \left(x\right)}\le {(1+\epsilon )}^{l}exp(-l{r}_{+}).\end{array}$$

Ultimately,

$$\begin{array}{cc}\hfill \frac{\phi (x-y)}{\phi \left(x\right)}& \le {(1+\epsilon )}^{l+1}exp(-r_{+}(y-l))exp(-l{r}_{+})={(1+\epsilon )}^{l+1}exp(-r_{+}y)\hfill \\ \hfill & \le (1+\epsilon )exp\left(\beta y\right),y\in [l,l+1],l=0,\cdots ,\left[x\right]-N-1.\hfill \end{array}$$

Now, let $y\in \left(\right[x]-N-1,x]$. We have

$$\frac{\phi (x-y)}{\phi \left(x\right)}\le \frac{\phi (N+2)}{\phi \left(x\right)}\le \frac{\phi (N+2)}{exp\left(r_{+}x\right)}\le \frac{\phi (N+2)}{exp\left(r_{+}y\right)}\le \phi (N+2)exp\left(\beta y\right).$$

Thus, the $U_{+}$-integrable majorant sought for the function $I(x,y)$, $y\in {\mathbb{R}}_{+}$, which does not depend on x, is of the form

$$\parallel U_{-}{\parallel }_{\phi }max\{(1+\epsilon ),\phi (N+2)\}exp\left(\beta y\right),y\in {\mathbb{R}}_{+}.$$

Now, in order to prove relation (19), it suffices, by Lebesgue’s theorem, to pass to the limit under the integral sign in (19). The last equality in (19) is a consequence of (8) for $\Re s=-r_{+}$:

$$\widehat{U}\left(s\right)=\frac{1}{1-\widehat{F}\left(s\right)}=\frac{1}{1-{\widehat{F}}_{-}\left(s\right)}\frac{1}{1-{\widehat{F}}_{+}\left(s\right)}={\widehat{U}}_{-}\left(s\right){\widehat{U}}_{+}\left(s\right),$$

which is admissible, since

$$|\widehat{F}\left(s\right)|\le \widehat{F}(-r_{+})<1,\left|{\widehat{F}}_{\pm }\left(s\right)\right|\le {\widehat{F}}_{\pm }(-r_{+})<1,\Re s=-r_{+}.$$

In the general case, it suffices to repeat the concluding reasoning of the previous proof using the estimate

$$\underset{x\to \infty }{lim\; sup}\frac{\left|z\right(x\left)\right|}{\phi \left(x\right)}\le \frac{C}{1-\widehat{F}(-r_{+})}$$

for $\left|g\right(x\left)\right|\le C\phi \left(x\right)$, $x\in {\mathbb{R}}_{+}$, and, considering the case $c=0$, take into account the relation $z_1\left(x\right)=o\left(x\right)$ as $x\to \infty$ and all the more $z_1\left(x\right)=o\left(\phi \left(x\right)\right)$ as $x\to \infty$, since $x\le e^{r_{+}x}\le \phi \left(x\right)$, $x\in {\mathbb{R}}_{+}$. □

4. Conclusions

We have established the asymptotic behavior of the solution z of the generalized Wiener–Hopf Equation (1), where the inhomogeneous term g behaves like an unbounded submultiplicative function, up to a constant factor, i.e., $g\left(x\right)\sim c\phi \left(x\right)$ as $x\to \infty$. Depending on whether $r_{+}=0$ or $r_{+}>0$, there are two different types of asymptotics for z (Theorems 2 and 3): either $z\left(x\right)\sim c_1{\int }_0^x\phi \left(y\right)dy$ or $z\left(x\right)\sim c_2\phi \left(x\right)$ as $x\to \infty$, where $c_1$ and $c_2$ are specific constants. Here are two simple examples ($c=1$):

(i) If $\phi \left(x\right)={(x+1)}^r$, $r>0$, then

$$z\left(x\right)\sim \frac{x^{r+1}}{\mu (r+1)}\mathrm{as}x\to \infty ;$$

(ii) If $\phi \left(x\right)=exp\left(\gamma x\right)$, $\gamma >0$, then

$$z\left(x\right)\sim \frac{e^{\gamma x}}{1-\widehat{F}(-r_{+})}\mathrm{as}x\to \infty .$$