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Article

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

by
Mikhail Sgibnev
Sobolev Institute of Mathematics, 630090 Novosibirsk, Russia
AppliedMath 2022, 2(3), 501-511; https://doi.org/10.3390/appliedmath2030029
Submission received: 17 August 2022 / Revised: 13 September 2022 / Accepted: 14 September 2022 / Published: 19 September 2022

## 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.
MSC:
45E10; 60K05

## 1. Introduction

The classical Wiener–Hopf equation has the form
$z ( x ) = ∫ 0 ∞ k ( x − y ) z ( y ) d y + g ( x ) , x ≥ 0 ,$
or, equivalently,
$z ( x ) = ∫ − ∞ x z ( x − y ) k ( y ) d y + g ( x ) , x ≥ 0 .$
We shall consider the inhomogeneous generalized Wiener–Hopf equation
$z ( x ) = ∫ − ∞ x z ( x − y ) F ( d y ) + g ( x ) , x ≥ 0 ,$
where z is the function sought, F is a given probability distribution on $R$, and the inhomogeneous term g is a known complex function. A probability distribution G on $R$ is called nonarithmetic if it is not concentrated on the set of points of the form 0, $± λ$, $± 2 λ$, … (see Section V.2, Definition 3 of [1]). Let $R +$ be the set of all nonnegative numbers and $R − : = R \ R +$ be the set of all negative numbers. For $c ∈ C$, we assume that $c / ∞$ is equal to zero. The relation $a ( x ) ∼ c b ( x )$ as $x → ∞$ means that $a ( x ) / b ( x ) → c$ as $x → ∞$; if $c = 0$, then $a ( x ) = o ( b ( x ) )$.
Definition 1.
A positive function $φ ( x )$, $x ∈ R$, is called submultiplicative if it is finite, Borel measurable, and satisfies the conditions: $φ ( 0 ) = 1$, $φ ( x + y ) ≤ φ ( x ) φ ( y )$, x, $y ∈ R$.
The following properties are valid for submultiplicative functions defined on the whole line (Theorem 7.6.2) of [2]:
$− ∞ < r − : = lim x → − ∞ log φ ( x ) x = sup x < 0 log φ ( x ) x ≤ inf x > 0 log φ ( x ) x = lim x → ∞ log φ ( x ) x = : r + < ∞ .$
Here are some examples of submultiplicative function on $R +$: (i) $φ ( x ) = ( x + 1 ) r$, $r > 0$; (ii) $φ ( x ) = exp ( c x β )$, where $c > 0$ and $0 < β < 1$; and (iii) $φ ( x ) = exp ( γ x )$, where $γ ∈ R$. In (i) and (ii), $r + = 0$, while in (iii), $r + = γ$. 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 $μ : = ∫ R x F ( d x )$ and the function $g ( x )$ is asymptotically equivalent (up to a constant factor) to a nondecreasing submultiplicative function $φ ( x )$ tending to infinity as $x → ∞$: $g ( x ) ∼ c φ ( x )$ as $x → ∞$. In the main theorems (Theorems 2 and 3), $φ ( x )$, $x ∈ R +$, is a nondecreasing submultiplicative function for which there exists $lim x → ∞ φ ( x + y ) / φ ( x )$ for each $y ∈ R$. If such a limit exists, then it is equal to $exp ( r + y )$.
Earlier [3], the asymptotic behavior of z was studied in detail under the following assumptions: (i) $μ ∈ ( 0 , + ∞ ]$ and (ii) g belong to either $g ∈ L 1 ( 0 , ∞ )$ or $g ∈ L ∞ ( 0 , ∞ )$. Roughly speaking, if $g ∈ L 1 ( 0 , ∞ )$, then $z ( x )$ tends to a specific finite limit as $x → ∞$. Moreover, under appropriate conditions, a submultiplicative rate of convergence was given in the form $o ( 1 / φ ( x ) )$. If $g ∈ L ∞ ( 0 , ∞ )$, then $z ( x ) = O ( x )$ or even $z ( x ) = f ( ∞ ) x / μ$ as $x → ∞$, provided $f ( ∞ ) : = lim x → ∞ f ( x )$ exists.
The existence of the solution to Equation (1) and its explicit form (5) were established in [4] for $g ∈ L ∞ ( 0 , ∞ )$ and arbitrary probability distributions F, regardless of whether F is of oscillating or drifting type. If $μ = 0$ and if some other hypotheses are fulfilled, then $z ( x )$ tends to a specific finite limit as $x → ∞$ (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 $S ( φ )$ of all complex-valued measures $ϰ$, such that
$∥ ϰ ∥ φ : = ∫ R φ ( x ) | ϰ | ( d x ) < ∞ ;$
here, $| ϰ |$ stands for the total variation of $ϰ$. The collection $S ( φ )$ is a Banach algebra with norm $∥ · ∥ φ$ by the usual operations of addition and scalar multiplication of measures; the product of two elements $ν$ and $ϰ$ of $S ( φ )$ is defined as their convolution $ν ∗ ϰ$ (Section 4.16) of [2]. The unit element of $S ( φ )$ is the measure $δ 0$ of unit mass concentrated at zero. Define the Laplace transform of a measure $ϰ$ as $ϰ ^ ( s ) : = ∫ R exp ( s x ) ϰ ( d x )$. It follows from (2) that the Laplace transform of any $ϰ ∈ S ( φ )$ converges absolutely with respect to $| ϰ |$ for all s in the strip $Π ( r − , r + ) : = { s ∈ C : r − ≤ ℜ s ≤ r + }$. Let $ν$ and $ϰ$ be two complex-valued measures on the $σ$-algebra $B$ of Borel sets in $R$. Their convolution is the measure
$ν ∗ ϰ ( A ) : = ∫ ∫ { x + y ∈ A } ν ( d x ) ϰ ( d y ) = ∫ R ν ( A − x ) ϰ ( d x ) , A ∈ B ,$
provided the integrals make sense; here, $A − x : = { y ∈ R : x + y ∈ A }$. Denote by $F n ∗$ the n-th convolution power of F:
$F 0 ∗ : = δ 0 , F 1 ∗ : = F , F ( n + 1 ) ∗ : = F n ∗ ∗ F , n ≥ 1 .$
Let U be the renewal measure generated by F: $U : = ∑ n = 0 ∞ F n ∗$.
Let $X k$, $k ≥ 1$, be independent random variables with the same distribution F not concentrated at zero. These variables generate the random walk $S 0 = 0$, $S n = X 1 + … + X n$, $n ≥ 1$. Put . The random variable $H ¯ + : = S T ¯ +$ is called the first weak ascending ladder height. Similarly, and $H ¯ − : = S T ¯ −$ is the first strong descending ladder height. We have the factorization identity (the symbol $E$ stands for “expectation”).
This can easily be deduced from an analogous identity in Section XVIII.3 of [1] for another collection of ladder variables. Denote by $F ±$ the distributions of the random variables $H ¯ ±$, respectively. It follows from the identity (3) that
$δ 0 − F = ( δ 0 − F − ) ∗ ( δ 0 − F + ) .$
Let $U ± : = ∑ k = 0 ∞ F ± k ∗$ be the renewal measures generated by the distributions $F ±$, respectively. Denote by $1 R +$ the indicator of the subset $R +$ in $R$: $1 R + ( x ) = 1$ for $x ∈ R +$ and $1 R + ( x ) = 0$ for $x ∈ R −$. Extend the function g onto the whole line: $g ( x ) : = 0$, $x < 0$. This convention will be valid throughout. Let $ν$ be a measure defined on $B$, and $a ( x )$, $x ∈ R$, a function. Define the convolution $ν ∗ a ( x )$ as the function $∫ R a ( x − y ) ν ( d y )$, $x ∈ R$. The following theorem has been proven in [4].
Theorem 1.
Let F be a probability distribution and $g ∈ L 1 ( R + )$. Then, the function
is the solution to Equation (1), which coincides with the solution obtained by successive approximations.
If $μ$ is finite and positive, then $μ + : = ∫ R x F + ( d x )$ is also finite and positive (Section XII.2, Theorem 2 of [1]). We have
$μ = μ + ( 1 − F − ( R − ) ) , U − ( R − ∪ { 0 } ) = 1 1 − F − ( R − ) .$
In fact, pass in (4) to Laplace transforms and divide both sides by s. We get
Let s tend to zero. Then, the fractions on both sides will tend to $μ$ and $μ +$, respectively. The second equality in (6) is a consequence of the fact that the distribution $F −$ is defective, i.e., $F − ( R − ) < 1$.
Lemma 1.
Let F be a nonarithmetic probability distribution, such that
$μ = ∫ R x F ( d x ) ∈ ( 0 , ∞ )$
and let $φ ( x )$, $x ∈ R$, be a submultiplicative function with $r − ≤ 0 ≤ r +$. Assume that
$∫ − ∞ 0 φ ( x ) F ( ( − ∞ , x ] ) d x < ∞ .$
Suppose additionally that $F ^ ( r − ) < 1$ if $r − < 0$. Then $U − ∈ S ( φ )$.
Proof.
By Theorem 4 in [7] with $n = 1$ and Remark 5 therein, we have
$∫ − ∞ 0 φ ( x ) F − ( d x ) < ∞ ,$
i.e., $F − ∈ S ( φ )$. Let us prove that the element $ν : = δ 0 − F −$ is invertible in $S ( φ )$. Let $ν = ν c + ν d + ν s$ be the decomposition of $ν$ into absolutely continuous, discrete, and singular components. By Theorem 1 of [8], the element $ν ∈ S ( φ )$ has an inverse if $ν ^ ( s ) ≠ 0$ for all $s ∈ Π ( r 1 , r 2 )$, and if
Let $F − = F − c + F − d + F − s$ be the decomposition of $F − ∈ S ( φ )$ into absolutely continuous, discrete, and singular components. Then, $ν d = δ 0 − F − d$ and $ν s = − F − s$. We have
On the other hand, . Hence, in order to prove (7), it suffices to show that
$1 − F − d ^ ( r − ) − F − s ^ ( r − ) ≥ 1 − F − ^ ( r − ) > 0 .$
If $r − = 0$, this follows from the fact that the distribution $F −$ is defective. Let $r − < 0$. By assumption, $F ^ ( r − ) < 1$ and, obviously, $F ^ + ( r − ) < 1$. Relation (4) implies
whence $1 − F − ^ ( r − ) > 0$ and (7) follows. Finally,
$| ν ^ ( s ) | ≥ 1 − | F − ^ ( s ) | ≥ 1 − F − ^ ( | s | ) ≥ 1 − F − ^ ( r − ) > 0 , s ∈ Π ( r − , r + ) .$
Therefore, by Theorem 1 in [8], the measure $δ 0 − F −$ is invertible in the Banach algebra $S ( φ )$ and $U − = ( δ 0 − F − ) − 1 ∈ S ( φ )$. The proof of the lemma is complete. □
Lemma 2.
Let $a ( x )$, $x ∈ R +$, be a monotone nondecreasing positive function. Suppose that $lim x → ∞ a ( x + y ) / a ( x ) = 1$ for each $y ∈ R$. Then,
Proof.
Let $M > 0$ be arbitrary. We have
$∫ 0 x a ( y ) a ( x ) d y ≥ ∫ x − M x a ( y ) a ( x ) d y ≥ ∫ x − M x a ( x − M ) a ( x ) d y = M a ( x − M ) a ( x ) .$
It follows that $lim inf x → ∞ ∫ 0 x a ( y ) d y / a ( x ) = ∞$. The proof of the lemma is complete. □
Lemma 3.
Let G be a nonarithmetic probability distribution on $R +$, such that
$μ G : = ∫ R x G ( d x ) ∈ ( 0 , ∞ )$
and let $U G$ be the corresponding renewal measure: $U G : = ∑ n = 0 ∞ G n ∗$. Suppose that $a ( x )$ and $b ( x )$, $x ∈ R +$, are nonnegative functions such that $a ( x ) ∼ b ( x )$ as $x → ∞$. Then,
$I ( x ) : = U G ∗ a ( x ) ∼ U G ∗ b ( x ) = : J ( x ) a s x → ∞ .$
Proof.
Given $ε > 0$, choose $A > 0$, such that
$( 1 − ε ) b ( x ) ≤ a ( x ) ≤ ( 1 + ε ) b ( x ) , x ≥ A .$
Let
Similarly, let $J ( x ) = J 1 ( x ) + J 2 ( x )$. Obviously,
$1 − ε ≤ lim inf x → ∞ I 1 ( x ) J 1 ( x ) ≤ lim sup x → ∞ I 1 ( x ) J 1 ( x ) ≤ 1 + ε .$
Since $ε$ is arbitrary, $lim x → ∞ I 1 ( x ) / J 1 ( x ) = 1$, i.e., $I 1 ( x ) ∼ J 1 ( x )$ as $x → ∞$. Moreover, $I 1 ( x ) ≥ a ( x − A ) U G ( [ 0 , x − A ] ) → ∞$ as $x → ∞$ by the elementary renewal theorem for the measure $U G$: $U G ( [ 0 , x ] ) ∼ x / μ G$ as $x → ∞$ (see Section 1.2 of [9]). According to Blackwell’s theorem (Section XI.1, Theorem 1 of [1]),
$I 2 ( x ) ≤ a ( A ) U G ( ( x − A , x ] ) → a ( A ) A / μ G as x → ∞ .$
Hence, $I ( x ) ∼ I 1 ( x )$ as $x → ∞$. A similar relation also holds for $J ( x )$, which completes the proof of the lemma. □
Lemma 4.
Let $φ ( x )$, $x ∈ R +$, be a submultiplicative function, such that there exists $ν ( y ) : = lim x → ∞ φ ( x + y ) / φ ( x )$ for each $y ∈ R$. Then $ν ( y ) = exp ( r + y )$, $y ∈ R$.
Proof.
By the Corollary of Theorem 4.17.3 in Section 4.17 of [2], $ν ( y ) = exp ( α y )$ for some $α ∈ R$. Given $ε > 0$, there exists $n 0 = n 0 ( ε )$, such that $log φ ( n + 1 ) φ ( n ) ≤ α + ε$ for $n ≥ n 0$.
Hence, $φ ( n 0 + m ) ≤ φ ( n 0 ) e m ( α + ε )$ and
$r + = lim m → ∞ log φ ( n 0 + m ) m ≤ lim m → ∞ log φ ( n 0 ) m + lim m → ∞ m ( α + ε ) m = α + ε .$
Similarly, $r + ≥ α − ε$. Since $ε > 0$ is arbitrary, $α = r +$. The proof of the lemma is complete. □

## 3. Main Results

Theorem 2.
Let F be a nonarithmetic probability distribution, such that
$μ = ∫ R x F ( d x ) ∈ ( 0 , ∞ )$
and let $φ ( x )$, $x ∈ R +$, be a nondecreasing continuous submultiplicative function tending to infinity as $x → ∞$, such that $r + = 0$ and there exists $lim x → ∞ φ ( x + y ) / φ ( x )$ for each $y ∈ R$. Suppose that the inhomogeneous term $g ( x )$, $x ∈ R +$, is bounded on finite intervals and satisfies the relation $g ( x ) ∼ c φ ( x )$ as $x → ∞$, where $c ∈ C$. Assume that
$∫ − ∞ 0 φ ( | x | ) F ( ( − ∞ , x ] ) d x < ∞ .$
Then, the function $z ( x )$, $x ∈ R +$, defined by (5) is a solution to Equation (1) and satisfies the asymptotic relation
$z ( x ) ∼ c μ ∫ 0 x φ ( y ) d y a s x → ∞ .$
Proof.
Put $M ( x ) = ∫ 0 x φ ( y ) d y$. By Lemma 4, $lim x → ∞ φ ( x + y ) / φ ( x ) = 1$ for each $y ∈ R$. Extend the function $φ ( x )$ onto the whole line $R$ by setting $φ ( x ) = φ ( | x | )$ for $x ∈ R −$. The extended function retains the submultiplicative property and $r ± = 0$. To prove the first statement of the theorem, it suffices to assume $g ≥ 0$. Choose $C > 0$, such that $g ( x ) ≤ C φ ( x )$, $x ∈ R +$. The function $z ( x )$ defined by (5) is finite, since
$U − ∗ g ( x ) ≤ C U − ∗ φ ( x ) = C ∫ − ∞ 0 + φ ( x − y ) U − ( d y ) ≤ C φ ( x ) ∥ U − ∥ φ , z ( x ) ≤ C ∥ U − ∥ φ ∫ 0 x φ ( x − y ) U + ( d y ) ≤ C ∥ U − ∥ φ φ ( x ) U + ( [ 0 , x ] ) < ∞$
for all $x ∈ R +$. Let n be a natural number. Denote by $1 [ 0 , n ]$ the indicator of $[ 0 , n ]$. Consider Equation (1) with the inhomogeneous term $g n ( x ) = g ( x ) 1 [ 0 , n ] ( x )$. Let $z n$ be the solution to the equation
$z n ( x ) = ∫ − ∞ x z n ( x − y ) F ( d y ) + g n ( x ) , x ∈ R + ,$
defined by formula (5):
The integral in (9) can be written as
$∫ R z n ( x − y ) 1 [ 0 , x ] ( y ) F ( d y ) ≤ z n ( x ) ≤ z ( x ) < ∞ .$
The last two inequalities are consequences of (5). Obviously, $z n ( x ) ↑$ as $n ↑$. By Section 27, Theorem B of [10], the integral tends to $∫ − ∞ x z ( x − y ) F ( d y )$ as $n ↑ ∞$. Letting $n ↑ ∞$ in (9) and (10), we get that z is a solution to (1). Let us prove the assertion of the theorem for the solution $z φ$ to (1) for $g = φ$. Let us show that
$U − ∗ φ ( x ) φ ( x ) → U − ( R − ∪ { 0 } ) as x → ∞ .$
We have
$U − ∗ φ ( x ) φ ( x ) = ∫ − ∞ 0 φ ( x − y ) φ ( x ) U − ( d y ) .$
By Lemma 4, the integrand tends to 1 as $x → ∞$ and it is majorized by the $U −$-integrable function $φ ( y )$, since
$φ ( x − y ) φ ( x ) ≤ φ ( − y ) = φ ( y )$
and $U − ∈ S ( φ )$ 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, $a ( x )$ and $b ( x )$:
$G : = F + , a ( x ) : = 1 R + ( x ) U − ∗ φ ( x ) , b ( x ) : = U − ( R − ∪ { 0 } ) 1 R + ( x ) φ ( x ) .$
We get
$z φ ( x ) = ∫ 0 x U − ∗ φ ( x − y ) U + ( d y ) ∼ U − ( R − ∪ { 0 } ) ∫ 0 x φ ( x − y ) U + ( d y ) as x → ∞ .$
Recalling (6), we see that in order to prove the theorem for $z φ$, it suffices to establish
$U + ∗ ( 1 R + φ ) ( x ) = ∫ 0 x φ ( x − y ) U + ( d y ) ∼ 1 μ + ∫ 0 x φ ( y ) d y = 1 μ + M ( x ) as x → ∞ .$
Integrating by parts, we get
$∫ 0 x φ ( x − y ) U + ( d y ) = φ ( x − y ) U + ( [ 0 , y ] ) | y = 0 x − ∫ 0 x U + ( [ 0 , y ] ) d y φ ( x − y ) = U + ( [ 0 , x ] ) − φ ( x ) − ∫ 0 x U + ( [ 0 , y ] ) d y φ ( x − y ) .$
The following three estimates hold:
$φ ( x ) , x , U + ( [ 0 , x ] ) = o ( M ( x ) ) as x → ∞ .$
The first estimate follows from Lemma 2 with $a ( x ) = φ ( x )$. The second one follows from the assumption $φ ( y ) → ∞$ as $y → ∞$. The third estimate follows from the second one and the elementary renewal theorem for the measure $U +$: $U + ( [ 0 , x ] ) ∼ x / μ +$ as $x → ∞$.
Show that
$− ∫ 0 x U + ( [ 0 , y ] ) d y φ ( x − y ) ∼ − 1 μ + ∫ 0 x y d y φ ( x − y ) as x → ∞ ,$
$− 1 μ + ∫ 0 x y d y φ ( x − y ) ∼ 1 μ + M ( x ) as x → ∞ .$
We prove first (17). This follows from the second estimate in (15) and the equality
$− ∫ 0 x y d y φ ( x − y ) = − y φ ( x − y ) | y = 0 x + ∫ 0 x φ ( x − y ) d y = − x + M ( x ) .$
Let $ε > 0$ be arbitrary. Use the elementary renewal theorem and choose $y 0 = y 0 ( ε )$, such that
$( 1 − ε ) U + ( [ 0 , y ] ) ≤ y μ + ≤ ( 1 + ε ) U + ( [ 0 , y ] ) , y ≥ y 0 .$
Write the left-hand side of (16) in the form
and let $M 1 ( x ) + M 2 ( x )$ be a similar decomposition for the right-hand side. Obviously,
$( 1 − ε ) M 2 ( x ) ≤ K 2 ( x ) ≤ ( 1 + ε ) M 2 ( x ) .$
Let us prove that, as $x → ∞$, both sides in (16) are asymptotically equivalent to $K 2 ( x )$ and $M 2 ( x )$, respectively. We have
$M 2 ( x ) = − 1 μ + ∫ y 0 x y d y φ ( x − y ) = − y μ + φ ( x − y ) | y = y 0 x + 1 μ + ∫ y 0 x φ ( x − y ) d y = − x μ + + y 0 μ + φ ( x − y 0 ) + 1 μ + ∫ 0 x − y 0 φ ( y ) d y .$
Let us show that
$M 3 ( x ) : = ∫ 0 x − y 0 φ ( y ) d y ∼ M ( x ) as x → ∞ .$
Using the first estimate in (15), we get
$∫ x − y 0 x φ ( y ) d y ≤ φ ( x ) ∫ x − y 0 x φ ( y − x ) d y = φ ( x ) ∫ 0 y 0 φ ( y ) d y ≤ φ ( x ) φ ( y 0 ) y 0 = o ( M ( x ) ) as x → ∞ .$
Finally,
which establishes the desired equivalence $M 3 ( x ) ∼ M ( x )$ as $x → ∞$. Taking into account the estimates in (15), we see that $M 2 ( x ) ∼ M ( x ) / μ +$ as $x → ∞$. Moreover,
$M 1 ( x ) = − − y 0 φ ( x − y 0 ) μ + + 1 μ + ∫ x − y 0 x φ ( u ) d u .$
The integral is estimated by $y 0 φ ( x ) / μ +$. Thus, $M 1 ( x ) = o ( M ( x ) )$ as $x → ∞$ (see (15)). Relation (17) is proven. Now, divide all parts of (18) by $M 2 ( x )$ and let x tend to infinity. We obtain
$1 − ε ≤ lim inf x → ∞ K 2 ( x ) M 2 ( x ) ≤ lim sup x → ∞ K 2 ( x ) M 2 ( x ) ≤ 1 + ε .$
Hence, $K 2 ( x ) ∼ M 2 ( x ) ∼ M ( x )$ as $x → ∞$. Relation (16) is proven, since, as $x → ∞$,
$K 1 ( x ) ≤ − U + ( [ 0 , y 0 ] ) ∫ 0 y 0 d y φ ( x − y ) = U + ( [ 0 , y 0 ] ) [ φ ( x ) − φ ( x − y 0 ) ] ≤ U + ( [ 0 , y 0 ] ) φ ( x ) = o ( M ( x ) ) .$
The equivalence (13) now follows from (14)–(17), which proves the theorem in the particular case $g = φ$. Let g satisfy the hypotheses of the theorem. If, for some $C > 0$, $| g ( x ) | ≤ C φ ( x )$, $x ∈ R +$, then
$lim sup x → ∞ | z ( x ) | / ∫ 0 x φ ( y ) d y ≤ C μ .$
It follows that if $c = 0$, then $z ( x ) = o ( z φ ( x ) )$ as $x → ∞$. To see this, choose a small $ε > 0$ and a natural number n, such that $| g ( x ) | ≤ ε φ ( x )$, $x ≥ n$. Write
$g = 1 [ 0 , n ] g + ( g − 1 [ 0 , n ] g ) = : g 1 + g 2 .$
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 $| z 2 ( x ) | ≤ ε z φ ( x )$, $x ∈ R +$. By Theorem 6.2 in [3], $z 1 ( x ) = o ( x )$ as $x → ∞$. Since $φ ( x ) ≥ 1$, $x ∈ R +$, it follows that as $x → ∞$. Therefore,
$lim sup x → ∞ | z ( x ) | / ∫ 0 x φ ( y ) d y ≤ ε μ .$
Since $ε > 0$ is arbitrary, the assertion of the theorem is true for $c = 0$. Let $c ≠ 0$. Write g in the form $g = c φ + g 1$. Then, $g 1 ( x ) = o ( φ ( x ) )$ as $x → ∞$, and we have $z = c z φ + z 1$, 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
$μ = ∫ R x F ( d x ) ∈ ( 0 , ∞ ) ,$
and let $φ ( x )$, $x ∈ R +$, be a nondecreasing submultiplicative function, such that $r + > 0$, and there exists $lim x → ∞ φ ( x + y ) / φ ( x )$ for each $y ∈ R$. Suppose that the inhomogeneous term $g ( x )$, $x ∈ R +$, is bounded on finite intervals and satisfies the relation $g ( x ) ∼ c φ ( x )$ as $x → ∞$, where $c ∈ C$. Assume that
$∫ − ∞ 0 φ ( | x | ) F ( ( − ∞ , x ] ) d x < ∞$
and $F ^ ( − r + ) < 1$. Then, the function $z ( x )$, $x ∈ R +$, defined by (5) is a solution to Equation (1) and satisfies the asymptotic relation
$z ( x ) ∼ c 1 − F ^ ( − r + ) φ ( x ) a s x → ∞ .$
Proof.
As in the proof of the preceding theorem, we verify that $z ( x )$ is a solution to (1). First, let us prove the assertion of the theorem for the solution $z φ$ to (1) corresponding to $g = φ$, i.e., let us prove that, as $x → ∞$,
$z φ ( x ) φ ( x ) = ∫ 0 x U − ∗ φ ( x − y ) φ ( x ) U + ( d y ) → U ^ − ( − r + ) U ^ + ( − r + ) = 1 1 − F ^ ( − r + ) .$
Write the integrand in the form
$I ( x , y ) : = 1 [ 0 , x ] ( y ) U − ∗ φ ( x − y ) φ ( x − y ) φ ( x − y ) φ ( x ) , y ∈ R + .$
Notice that
$U − ∗ φ ( x ) φ ( x ) = ∫ − ∞ 0 φ ( x − y ) φ ( x ) U − ( d y ) → U ^ − ( − r + ) as x → ∞ .$
In fact, $φ ( x − y ) / φ ( x ) → e − r + y$ as $x → ∞$ by Lemma 4 and, according to Lemma 1, this ratio is majorized by the $U −$-integrable function $φ ( y )$, $y ∈ R −$:
$U − ∗ φ ( x ) φ ( x ) = ∫ − ∞ 0 φ ( x − y ) φ ( x ) U − ( d y ) ≤ ∫ − ∞ 0 φ ( | y | ) U − ( d y ) = ∥ U − ∥ φ < ∞ .$
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 $I ( x , y )$, $y ∈ R +$, in the form $M e β y$ with $β ∈ ( − r + , 0 )$. Then, by Lebesgue’s theorem, we pass to the limit under the integral sign in the left-side integral in (19) as $x → ∞$, and thus prove relation (19). Put $f ( x ) = log φ ( x ) − r + x$. By hypothesis, we have
$f ( x − y ) − f ( x ) = log φ ( x − y ) − log φ ( x ) + r + x → 0 as x → ∞$
for each $y ∈ R$. According to Lemma 1.1 in [11], relation (21) is fulfilled uniformly in $y ∈ [ 0 , 1 ]$. Hence,
$φ ( x − y ) exp ( r + y ) φ ( x ) → 1 as x → ∞$
uniformly in $y ∈ [ 0 , 1 ]$. Choose a small $ε > 0$ such that $β : = log ( 1 + ε ) − r + < 0$. Let $N = N ( ε ) > 0$ be an integer such that
$φ ( x − y ) exp ( r + y ) φ ( x ) ≤ 1 + ε , x ≥ N , y ∈ [ 0 , 1 ] .$
Denote by $[ x ]$ the integral part of a real number x; i.e., $[ x ]$ is the maximal integer not exceeding x: $x = [ x ] + ϑ$, $ϑ ∈ [ 0 , 1 )$. For $y ∈ [ l , l + 1 ]$, $l = 0 , … , [ x ] − N − 1$, we have
$φ ( x − y ) φ ( x ) = φ ( x − l − ( y − l ) ) φ ( x − l ) φ ( x − l ) φ ( x ) , φ ( x − l − ( y − l ) ) φ ( x − l ) ≤ ( 1 + ε ) exp ( − r + ( y − l ) ) , φ ( x − l ) φ ( x ) = φ ( x − l ) φ ( x − l + 1 ) φ ( x − l + 1 ) φ ( x − l + 2 ) … φ ( x − 1 ) φ ( x ) ≤ ( 1 + ε ) l exp ( − l r + ) .$
Ultimately,
$φ ( x − y ) φ ( x ) ≤ ( 1 + ε ) l + 1 exp ( − r + ( y − l ) ) exp ( − l r + ) = ( 1 + ε ) l + 1 exp ( − r + y ) ≤ ( 1 + ε ) exp ( β y ) , y ∈ [ l , l + 1 ] , l = 0 , ⋯ , [ x ] − N − 1 .$
Now, let $y ∈ ( [ x ] − N − 1 , x ]$. We have
$φ ( x − y ) φ ( x ) ≤ φ ( N + 2 ) φ ( x ) ≤ φ ( N + 2 ) exp ( r + x ) ≤ φ ( N + 2 ) exp ( r + y ) ≤ φ ( N + 2 ) exp ( β y ) .$
Thus, the $U +$-integrable majorant sought for the function $I ( x , y )$, $y ∈ R +$, which does not depend on x, is of the form
$∥ U − ∥ φ max { ( 1 + ε ) , φ ( N + 2 ) } exp ( β y ) , y ∈ R + .$
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 $ℜ s = − r +$:
$U ^ ( s ) = 1 1 − F ^ ( s ) = 1 1 − F ^ − ( s ) 1 1 − F ^ + ( s ) = U ^ − ( s ) U ^ + ( s ) ,$
$| F ^ ( s ) | ≤ F ^ ( − r + ) < 1 , | F ^ ± ( s ) | ≤ F ^ ± ( − r + ) < 1 , ℜ s = − r + .$
In the general case, it suffices to repeat the concluding reasoning of the previous proof using the estimate
$lim sup x → ∞ | z ( x ) | φ ( x ) ≤ C 1 − F ^ ( − r + )$
for $| g ( x ) | ≤ C φ ( x )$, $x ∈ R +$, and, considering the case $c = 0$, take into account the relation $z 1 ( x ) = o ( x )$ as $x → ∞$ and all the more $z 1 ( x ) = o ( φ ( x ) )$ as $x → ∞$, since $x ≤ e r + x ≤ φ ( x )$, $x ∈ R +$. □

## 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., $g ( x ) ∼ c φ ( x )$ as $x → ∞$. Depending on whether $r + = 0$ or $r + > 0$, there are two different types of asymptotics for z (Theorems 2 and 3): either $z ( x ) ∼ c 1 ∫ 0 x φ ( y ) d y$ or $z ( x ) ∼ c 2 φ ( x )$ as $x → ∞$, where $c 1$ and $c 2$ are specific constants. Here are two simple examples ($c = 1$):
(i) If $φ ( x ) = ( x + 1 ) r$, $r > 0$, then
$z ( x ) ∼ x r + 1 μ ( r + 1 ) as x → ∞ ;$
(ii) If $φ ( x ) = exp ( γ x )$, $γ > 0$, then
$z ( x ) ∼ e γ x 1 − F ^ ( − r + ) as x → ∞ .$

## Funding

This research received no external funding.

Not applicable.

Not applicable.

Not applicable.

## Acknowledgments

The work was carried out within the framework of the State Task to the Sobolev Institute of Mathematics (Project FWNF-2022-0004).

## Conflicts of Interest

The author declares no conflict of interest.

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Sgibnev, M. The Wiener–Hopf Equation with Probability Kernel and Submultiplicative Asymptotics of the Inhomogeneous Term. AppliedMath 2022, 2, 501-511. https://doi.org/10.3390/appliedmath2030029

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Sgibnev M. The Wiener–Hopf Equation with Probability Kernel and Submultiplicative Asymptotics of the Inhomogeneous Term. AppliedMath. 2022; 2(3):501-511. https://doi.org/10.3390/appliedmath2030029

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Sgibnev, Mikhail. 2022. "The Wiener–Hopf Equation with Probability Kernel and Submultiplicative Asymptotics of the Inhomogeneous Term" AppliedMath 2, no. 3: 501-511. https://doi.org/10.3390/appliedmath2030029