A Tauberian Theorem of Wiener–Ikehara Type and Its Applications

Abstract

In this paper, we establish a Tauberian theorem of Wiener–Ikehara type. As an application, we give a short proof of Blackwell’s renewal theorem of discrete case. We also establish a Dirichlet series version of Blackwell’s renewal theorem as another application.

1 Introduction

Let $A\subset {\mathbb {R}}$ and $Y\subset {\mathbb {C}}$ be measurable. Let $L^1(A;Y)$ be defined as the set of all measurable $f:A\rightarrow Y$ such that $\int _A|f(t)|dt<+\infty$. Let $L^1_{loc}(A;Y)$ be defined as the set of all measurable $f:A\rightarrow Y$ such that $\int _K|f(t)|dt<+\infty$ for any compact $K\subset {\mathbb {R}}$ with $K\subset A$. We also define $L^p(A;Y)$ and $L^p_{loc}(A;Y)$ $(1<p\le +\infty )$ in a similar manner. In this paper, we are concerned with Tauberian theorems of Wiener–Ikehara type. One of a classical Wiener–Ikehara theorem can be stated as follows:

Theorem 1

Let $f\in L^1_{loc}({\mathbb {R}_{\ge 0}};{\mathbb {R}})$ be a function such that $f(x)e^x~(x\ge 0)$ is non-decreasing and the Lebesgue integral $F(s):=\int _0^\infty f(x)e^{-sx}dx$ converges in ${\mathbb {C}}$ for $\mathfrak {R}(s)>0$. If there exists $A\in {\mathbb {R}}$ and $B\in L^1_{loc}({\mathbb {R}};{\mathbb {C}})$ such that $\lim _{\sigma \downarrow 0}\int _{-\delta }^\delta |F(\sigma +it)-A/(\sigma +it)-B(t)|dt=0~(\forall \delta >0)$, then we have $\lim _{x\rightarrow +\infty }f(x)=A$.

Today, several generalizations of the Wiener–Ikehara theorem have been given. One possible generalization is to replace the non-decreasing condition of f with a weaker condition. In this direction, Zhang [12] established an exact form (i.e., necessary and sufficient form) of the Wiener–Ikehara theorem by using the concept of log-linearly slowly decreasing condition. Another generalization is to replace the boundary behavior $\lim _{\sigma \downarrow 1}\int _{-\delta }^\delta |F(\sigma +it)-A/(\sigma +it)-B(t)|dt=0$ with a weaker condition. In this direction, Korevaar [9] established another exact form of the Wiener–Ikehara theorem by using the concept of local pseudofunction boundary behavior. Moreover, those two generalizations can be combined (e.g., [3]). We refer to [4, 5, 10, 11] for further developments around the Wiener–Ikehara theorem.

In this paper, we are mainly concerned with boundary behavior, and we provide sufficient conditions which guarantee the existence of $\lim _{x\rightarrow +\infty }f(x)$. Since our conditions are relatively easy to check, we will provide some applications later.

Our main theorem is the following:

Theorem 2

Let $f\in L^1_{loc}({\mathbb {R}_{\ge 0}};{\mathbb {R}})$ satisfy $\int _0^\infty |f(x)|/(x+1)^2dx<+\infty$. In addition, assume that either we have

$$\begin{aligned}&\forall \varepsilon>0,~\exists h>0,~\exists u_0>0,~\forall u,v>0~ \nonumber \\&\left[ ~ u_0\le u\le v\le u+h~\Rightarrow ~ e^vf(v)-e^uf(u)\ge -\varepsilon e^u ~\right] , \end{aligned}$$

(1)

or we have

$$\begin{aligned}&\forall \varepsilon>0,~\exists h>0,~\exists u_0>0,~\forall u,v>0~ \nonumber \\&\left[ ~ u_0\le u\le v\le u+h~\Rightarrow ~ e^vf(v)-e^uf(u)\le \varepsilon e^u ~\right] . \end{aligned}$$

(2)

Since $\int _0^\infty |f(x)|/(x+1)^2dx<+\infty$, we can define $F:\left\{ s\in {\mathbb {C}}\mid \mathfrak {R}(s)>0 \right\} \rightarrow {\mathbb {C}}$ as $F(s):=\int _0^\infty f(x)e^{-sx}dx$. Assume that there exists $\left\{ \sigma _n \right\} _{n=0}^\infty \subset {\mathbb {R}}_{>0}$ with $\lim _n \sigma _n=0$ such that

• (i) $\forall u,v\in {\mathbb {R}}_{>0}~\left[ ~ 0 <u\le v~\Rightarrow ~ \lim _N\sup _{n,m\ge N}\int _u^v|\mathfrak {R}F(\sigma _n+it)-\mathfrak {R}F(\sigma _m\right.$$\left. +it)|dt=0~\right]$,

• (ii) $\exists T>0,~\exists g\in L^1([0,T];{\mathbb {R}}),~\forall n\ge 0,~a.e.t\in [0,T]~\left[ ~ \mathfrak {R}F(\sigma _n+it)\ge g(t) ~\right]$.

Then $\lim _{x\rightarrow +\infty }f(x)$ exists in ${\mathbb {R}}$.

Since the corresponding result is also true for a certain type of sequences, we also state this result:

Theorem 3

Let $a:{\mathbb {Z}_{\ge 0}}\rightarrow {\mathbb {R}}$ satisfy $\sum _{n=0}^\infty |a_n|/(n+1)^2<+\infty$. Note that we can define $F:\left\{ z\in {\mathbb {C}}\mid |z|<1 \right\} \rightarrow {\mathbb {C}}$ as $F(z):=\sum _{n=0}^\infty a_nz^n$. Assume that there exists $\left\{ r_n \right\} _{n=0}^\infty \subset (0,1)$ with $\lim _n r_n=1$ such that

• (i) $\forall v\in (0,\pi ]~\left[ ~ \lim _N\sup _{n,m\ge N}\int _v^\pi |\mathfrak {R}F(r_ne^{it})-\mathfrak {R}F(r_me^{it})|dt=0 ~\right]$,

• (ii) $\exists T\in (0,\pi ],~\exists g\in L^1([0,T];{\mathbb {R}}),~\forall n\ge 0,~a.e.t\in [0,T]~\left[ ~ \mathfrak {R}F(r_ne^{it})\ge g(t) ~\right]$.

Then $\lim _{n\rightarrow +\infty }a_n$ exists in ${\mathbb {R}}$.

Remark 1

The Wiener–Ikehara theorem usually requires a constant $A\in {\mathbb {R}}$ with the term “$F(\sigma +it)-A/(\sigma +it)$”. This A is usually prepared to obtain $\lim _{x\rightarrow +\infty }f(x)=A$. On the other hand, Theorem 2 does not require such A. We will later prove that the whole conditions of Theorem 2 automatically provide an A. This result is partially due to the condition $\int _0^\infty |f(x)|/(x+1)^2dx<+\infty$, so we need this condition, while the Wiener–Ikehara theorem does not require it in general. Incidentally, we usually apply the Wiener–Ikehara theorem for only bounded functions, and in this case we trivially obtain $\int _0^\infty |f(x)|/(x+1)^2dx<+\infty$. Therefore, the condition $\int _0^\infty |f(x)|/(x+1)^2dx<+\infty$ does not restrict the use of Theorem 2 in practice.

Remark 2

The conditions (1) and (2) of Theorem 2 are equivalent to the log-linearly slowly decreasing (increasing) condition [3, 12], and this condition is not our main concern. The important conditions of Theorem 2 are of course (i) and (ii). The condition (i) states that $\mathfrak {R}F(\sigma _n+it)~(n\ge 0)$ is a cauchy sequence in $L^1([u,v];{\mathbb {R}})$ for any $0<u\le v$. Note that $u=0$ is excluded. The condition (ii) states the good boundary behavior of $\mathfrak {R}F(\sigma +it)$ in the neighborhood of the origin $0+i0$.

Remark 3

The assumptions of Theorem 1 imply that of Theorem 2, provided that $\int _0^\infty |f(x)|/(x+1)^2dx<+\infty$. Let us briefly verify this fact. Let f satisfy the assumptions of Theorem 1 and $\int _0^\infty |f(x)|/(x+1)^2dx<+\infty$. Since $f(x)e^x$ is non-decreasing, f satisfies (1). Next, we show that the condition $\lim _{\sigma \downarrow 0}\int _{-\delta }^\delta |F(\sigma +it)-A/(\sigma +it)-B(t)|dt=0~(\forall \delta >0)$ implies (i) and (ii) of Theorem 2. We first assume $A\ge 0$. Let $\varepsilon _\sigma (t):=F(\sigma +it)-A/(\sigma +it)-B(t)~(\sigma >0,~t\in {\mathbb {R}})$. By assumption, we have $\lim _{\sigma \downarrow 0}\int _{-\delta }^\delta |\varepsilon _\sigma (t)|dt=0~(\delta >0)$. In particular, $\lim _{\sigma \downarrow 0}\int _0^1|\varepsilon _\sigma (t)|dt=0$. Then there exists $\left\{ \sigma _n \right\} _{n=0}^\infty \subset {\mathbb {R}}_{>0}$ with $\lim _n \sigma _n=0$ such that $\int _0^1|\varepsilon _{\sigma _n}(t)|dt\le 1/2^n~(n\ge 0)$. If we define $h(t):=\sum _{k=0}^\infty |\varepsilon _{\sigma _k}(t)|\in [0,+\infty ]~(t\in [0,1])$, then $\int _0^1 h\le 2<+\infty$, so $0\le h(t)<+\infty ~(a.e.t\in [0,1])$. Now let $T:=1$ and $g(t):=-|B(t)|-h(t)~(a.e.t\in [0,T])$. Then $g\in L^1([0,T];{\mathbb {R}})$. Moreover, we can easily show (i) and (ii) of Theorem 2, where we use $\mathfrak {R}(A/(\sigma +it))=A\sigma /(\sigma ^2+t^2)\ge 0$ to obtain $\mathfrak {R}F(\sigma _n+it)\ge g(t)$ in (ii). If $A<0$, then $-f$ satisfies $\int _0^\infty |-f(x)|/(x+1)^2dx<+\infty$. Moreover, $-f$ satisfies (2) and $-A>0$, so we can apply the same argument.

Remark 4

In Theorem 3, we do not need the corresponding conditions of (1) and (2).

The rest of this paper is structured as follows. In Sect. 2, we provide some lemmas related to (classical) Tauberian methods. In Sect. 3, we prove some crucial lemmas. These lemmas enable us to provide an $A\in {\mathbb {R}}$ which satisfies $\lim _{x\rightarrow +\infty }f(x)=A$. In Sect. 4, we prove Theorem 2. In Sect. 5, we prove Theorem 3. In Sects. 6 and 7, we provide some applications.

2 Tauberian Methods

In this section, we prove some lemmas related to (classical) Tauberian methods.

Lemma 1

Let $f:{\mathbb {R}_{\ge 0}}\rightarrow {\mathbb {R}}$ satisfy (1). Then we have the following:

• (i) $\forall \varepsilon>0,~\exists h\in (0,\varepsilon ),~\exists u_0>0,~\forall u,v>0$

  $\left[ ~ u_0\le u\le v\le u+h~\Rightarrow ~ e^vf(v)-e^uf(u)\ge -\varepsilon e^u ~\right]$

• (ii) $\forall \varepsilon>0,~\exists h\in (0,\varepsilon ),~\exists u_0>0,~\forall u,v>0$

  $\left[ ~ u_0\le u-h\le v\le u+h~\Rightarrow ~ e^vf(v)-e^{u-h}f(u-h)\ge -\varepsilon e^{u-h} ~\right]$.

• (iii) $\forall \varepsilon>0,~\exists h\in (0,\varepsilon ),~\exists u_0>0,~\forall u,v>0$

  $\left[ ~ u_0\le v-h\le u\le v+h~\Rightarrow ~ e^{v+h}f(v+h)-e^uf(u)\ge -\varepsilon e^u ~\right]$.

Proof

It is easy to verify (i). Next, we show (ii). Let $\varepsilon >0$. We take $h\in (0,\varepsilon )$ and $u_0>0$ of (i). Let $h_1:=h/2\in (0,\varepsilon )$. If $u,v>0$ satisfy $u_0\le u-h_1\le v\le u+h_1$, then $u_0\le (u-h_1)\le v\le (u-h_1)+h$ with $u-h_1,v>0$, so $e^vf(v)-e^{(u-h_1)}f(u-h_1)\ge -\varepsilon e^{(u-h_1)}$. Thus we obtain (ii). We can also show (iii) in a similar manner. $\square$

Lemma 2

Let $f:{\mathbb {R}_{\ge 0}}\rightarrow {\mathbb {R}}$ satisfy (1). Then $\inf _{x\ge x_0}f(x)>-\infty$ for some $x_0\ge 0$.

Proof

This is essentially shown in the proof of [3, Prop. 3.1], see [3, Eq. (3.4)]. $\square$

Lemma 3

Let $g:{\mathbb {R}}\rightarrow {\mathbb {R}_{\ge 0}}$ satisfy $g\in L^1({\mathbb {R}};{\mathbb {R}_{\ge 0}})\cap L^\infty ({\mathbb {R}};{\mathbb {R}_{\ge 0}}),~\int _{-\infty }^\infty g(x)dx>0$, and $\lim _{x\rightarrow ~-\infty }g(x)=0$.

(i) Let $f\in L^1_{loc}({\mathbb {R}_{\ge 0}};{\mathbb {R}})$ satisfy (1). Assume that the Lebesgue integral $\int _0^\infty f(x)\lambda g(\lambda (x-y))dx$ converges in ${\mathbb {R}}$ for any $\lambda ,y>0$. Then we have the following:

$$\begin{aligned} (\limsup _{y\rightarrow +\infty }f(y))\int _{-\infty }^\infty g(x)dx \le \liminf _{\lambda \rightarrow +\infty }\limsup _{y\rightarrow +\infty }\int _0^\infty f(x)\lambda g(\lambda (x-y))dx. \end{aligned}$$

(3)

If in addition the right hand side of (3) is not $+\infty$, then $\sup _{x\ge x_1}f(x)<+\infty$ for some $x_1\ge 0$, and we have

$$\begin{aligned} (\liminf _{y\rightarrow +\infty }f(y))\int _{-\infty }^\infty g(x)dx \ge \limsup _{\lambda \rightarrow +\infty }\liminf _{y\rightarrow +\infty }\int _0^\infty f(x)\lambda g(\lambda (x-y))dx. \end{aligned}$$

(4)

(ii) Let $f\in L^1_{loc}({\mathbb {R}_{\ge 0}};{\mathbb {R}})$ satisfy (2). Assume that the Lebesgue integral $\int _0^\infty f(x)\lambda g(\lambda (x-y))dx$ converges in ${\mathbb {R}}$ for any $\lambda ,y>0$. Then we have (4). If in addition the right hand side of (4) is not $-\infty$, then $\inf _{x\ge x_1}f(x)>-\infty$ for some $x_1\ge 0$, and we have (3).

Proof

(i) By Lemma 2, we have $f(x)\ge -C~(x\ge x_0)$ for some $C>0$ and $x_0\ge 0$.

STEP1: This step is devoted to proving (3). Let $\varepsilon >0$. By Lemma 1, we can take $h\in (0,\varepsilon )$ and $u_0>0$ such that (ii) of Lemma 1 holds. For any $u_0\le y-h\le x\le y+h$ with $x,y>0$, we have $e^xf(x)-e^{y-h}f(y-h)\ge -\varepsilon e^{y-h}$, i.e., $f(x)\ge e^{y-x-h}(f(y-h)-\varepsilon )$. Next, by the dominated convergence theorem (DCT, for short), we have $\lim _{\lambda \rightarrow +\infty }\int _{-\lambda h}^{\lambda h} e^{-t/\lambda }g(t)dt=\int _{-\infty }^\infty g(t)dt>0$, so we have $\int _{-\lambda h}^{\lambda h} e^{-t/\lambda }g(t)dt>0~(\forall \lambda >\lambda _0)$ for some $\lambda _0>0$. Now let $\lambda >\lambda _0$ and $y>u_0+x_0+h$. Then

$$\begin{aligned}&\int _0^\infty f(x)\lambda g(\lambda (x-y))dx =\left( \int _0^{x_0}+\int _{x_0}^{y-h}+\int _{y-h}^{y+h}+\int _{y+h}^\infty \right) f(x)\lambda g(\lambda (x-y))dx \\&\ge \int _0^{x_0}f(x)\lambda g(\lambda (x-y))dx-C\left( \int _{x_0}^{y-h}+\int _{y+h}^\infty \right) \lambda g(\lambda (x-y))dx \\&+(f(y-h)-\varepsilon )\int _{y-h}^{y+h}e^{y-x-h}\lambda g(\lambda (x-y))dx \\&=\int _0^{x_0}f(x)\lambda g(\lambda (x-y))dx-C\left( \int _{-\lambda (y-x_0)}^{-\lambda h}+\int _{\lambda h}^\infty \right) g(t)dt \\&+(f(y-h)-\varepsilon )e^{-h}\int _{-\lambda h}^{\lambda h} e^{-t/\lambda }g(t)dt. \end{aligned}$$

Since $y>u_0+x_0+h$ is arbitrary, we take $\limsup _{y\rightarrow +\infty }$. In view of the assumption on f and g, we have $\lim _{y\rightarrow +\infty }\int _0^{x_0}f(x)\lambda g(\lambda (x-y))dx=0$ by the DCT. We also have $\lim _{y\rightarrow +\infty }\int _{-\lambda (y-x_0)}^{-\lambda h}g(t)dt=\int _{-\infty }^{-\lambda h}g(t)dt$. Then

$$\begin{aligned}&\limsup _{y\rightarrow +\infty }\int _0^\infty f(x)\lambda g(\lambda (x-y))dx \nonumber \\&\quad \ge -C\left( \int _{-\infty }^{-\lambda h}+\int _{\lambda h}^\infty \right) g(t)dt+ \limsup _{y\rightarrow +\infty }\left( (f(y-h)-\varepsilon )e^{-h}\int _{-\lambda h}^{\lambda h} e^{-t/\lambda }g(t)dt\right) \nonumber \\&\quad =-C\left( \int _{-\infty }^{-\lambda h}+\int _{\lambda h}^\infty \right) g(t)dt+ (\limsup _{y\rightarrow +\infty }f(y)-\varepsilon )e^{-h}\int _{-\lambda h}^{\lambda h} e^{-t/\lambda }g(t)dt. \end{aligned}$$

(5)

Since $f(x)\ge -C~(x\ge x_0)$, we have $\limsup _{y\rightarrow +\infty }f(y)\in (-\infty ,+\infty ]$. If $\limsup _{y\rightarrow +\infty }f(y)<+\infty$, then we have $\limsup _{y\rightarrow +\infty }f(y)\in {\mathbb {R}}$. Since (5) holds for any $\lambda >\lambda _0$, we take $\liminf _{\lambda \rightarrow +\infty }$ in (5). In view of the DCT, we obtain

$$\begin{aligned} \liminf _{\lambda \rightarrow +\infty }\limsup _{y\rightarrow +\infty }\int _{-\infty }^\infty f(x)\lambda g(\lambda (x-y))dx \ge (\limsup _{y\rightarrow +\infty }f(y)-\varepsilon )e^{-h}\int _{-\infty }^\infty g(t)dt. \end{aligned}$$

Since $\varepsilon >0$ is arbitrary, we take $\varepsilon \downarrow 0$. Although h depends on $\varepsilon$, we have $h\in (0,\varepsilon )$, so $h\rightarrow 0$, provided that $\varepsilon \downarrow 0$. Hence, we obtain (3). The remaining possibility is $\limsup _{y\rightarrow +\infty }f(y)=+\infty$. In this case, we have $\limsup _{y\rightarrow +\infty }\int _0^\infty f(x)\lambda g(\lambda (x-y))dx=+\infty$ by (5). Since $\lambda >\lambda _0$ is arbitrary, we have $\liminf _{\lambda \rightarrow +\infty }\limsup _{y\rightarrow +\infty }\int _0^\infty f(x)\lambda g(\lambda (x-y))dx=+\infty$, so we trivially obtain (3). Thus we complete the proof of (3).

STEP2: By STEP1, we now obtain (3). If in addition the right hand side of (3) is not $+\infty$, then it follows that $\limsup _{y\rightarrow +\infty }f(y)<+\infty$, so we have $\sup _{x\ge x_1}f(x)<+\infty$ for some $x_1\ge 0$. Then $f(x)\le C_1~(x\ge x_1)$ for some $C_1>0$. Now we show (4). Let $\varepsilon >0$. By Lemma 1, we can take $h\in (0,\varepsilon )$ and $u_1>0$ such that (iii) of Lemma 1 holds. For any $u_1\le y-h\le x\le y+h$ with $x,y>0$, we have $e^{y+h}f(y+h)-e^xf(x)\ge -\varepsilon e^x$. In particular, $e^xf(x)\le e^{y+h}f(y+h)+\varepsilon e^x\le e^{y+h}f(y+h)+\varepsilon e^{y+h}=e^{y+h}(f(y+h)+\varepsilon )$, i.e., $f(x)\le e^{y-x+h}(f(y+h)+\varepsilon )$. Next, we have $\int _{-\lambda h}^{\lambda h} e^{-t/\lambda }g(t)dt>0~(\forall \lambda >\lambda _1)$ for some $\lambda _1>0$. Now we can show (4) by an argument similar to the above.

(ii) We can apply (i) for $-f$, and we complete the proof. $\square$

Corollary 1

Let $g:{\mathbb {R}}\rightarrow {\mathbb {R}_{\ge 0}}$ satisfy $g\in L^1({\mathbb {R}};{\mathbb {R}_{\ge 0}})\cap L^\infty ({\mathbb {R}};{\mathbb {R}_{\ge 0}}),~\int _{-\infty }^\infty g(x)dx>0$, and $\lim _{x\rightarrow ~-\infty }g(x)=0$. Let $f\in L^1_{loc}({\mathbb {R}_{\ge 0}};{\mathbb {R}})$ satisfy (1) or (2). Assume that the Lebesgue integral $\int _0^\infty f(x)\lambda g(\lambda (x-y))dx$ converges in ${\mathbb {R}}$ for any $\lambda ,y>0$. In addition, assume that

$$\begin{aligned} \exists C\in {\mathbb {R}},~\forall \lambda >0~\left[ ~ \lim _{y\rightarrow +\infty }\int _0^\infty f(x)\lambda g(\lambda (x-y))dx=C ~\right] . \end{aligned}$$

Then $\lim _{x\rightarrow +\infty }f(x)$ exists in ${\mathbb {R}}$, and in fact $\lim _{x\rightarrow +\infty }f(x)=C/\int _{-\infty }^\infty g(x)dx$.

Remark 5

The conditions (1) and (2) are equivalent to log-linearly slowly decreasing (increasing) condition [3, 12], so we can show that Corollary 1 is an exact form: Let $f\in L^1_{loc}({\mathbb {R}_{\ge 0}};{\mathbb {R}})$ satisfy $\lim _{x\rightarrow +\infty }f(x)=\alpha \in {\mathbb {R}}$. Then we can easily show that f satisfies (1) and (2). Moreover, the Lebesgue integral $\int _0^\infty f(x)\lambda g(\lambda (x-y))dx$ converges in ${\mathbb {R}}$ for any $\lambda ,y>0$. Furthermore, we can show $\lim _{y\rightarrow +\infty }\int _0^\infty f(x)\lambda g(\lambda (x-y))dx=\alpha \int _{-\infty }^\infty g(x)dx$ for any $\lambda >0$. Therefore, Corollary 1 can be regarded as a necessary and sufficient form, while this topic is not our main concern.

3 Crucial Lemmas

In this section, we provide some crucial lemmas.

Definition 1

For $T>0$, we define $I([0,T];{\mathbb {R}}):=\{a\in L^\infty ([0,T];{\mathbb {R}})\mid \lim _{t\downarrow 0}a(t)=a(0)\}$. We also define $I({\mathbb {R}_{\ge 0}};{\mathbb {R}}):=\{a\in L^\infty _{loc}({\mathbb {R}_{\ge 0}};{\mathbb {R}})\mid \lim _{t\downarrow 0}a(t)=a(0)\}$.

Lemma 4

Let $0<T\le U$. Let $\left\{ f_n \right\} _{n\ge 0}\subset L^1([0,U];{\mathbb {R}})$. Let $B:[0,U]\rightarrow {\mathbb {R}}$ satisfy $B\in L^1([v,U];{\mathbb {R}})~(\forall v\in (0,U])$. Assume that

• $\forall v\in (0,U]~\left[ ~ \lim _n \int _v^U|f_n(t)-B(t)|dt=0 ~\right]$,

• $\forall v\in (0,T]~\left[ ~ \lim _n (1/v)\int _0^v(v-t)f_n(t)dt~exists~in~{\mathbb {R}} ~\right]$,

• $\exists g\in L^1([0,T];{\mathbb {R}}),~\forall n\ge 0,~a.e.t\in [0,T]~\left[ ~ f_n(t)\ge g(t) ~\right]$.

Then we have $B\in L^1([0,U];{\mathbb {R}})$. Moreover, there exists $C\in {\mathbb {R}}$ such that

$$\begin{aligned} \forall a\in I([0,U];{\mathbb {R}}),~\forall v\in (0,U]~\left[ ~ \lim _n \int _0^va(t)f_n(t)dt=\int _0^va(t)B(t)dt+Ca(0) ~\right] . \end{aligned}$$

Proof

We first deal with the case $g(t)=0~(t\in [0,T])$, and we prove the lemma in this case. Note that we have $f_n(t)\ge 0~(n\ge 0,~a.e.t\in [0,T])$. Let $F:(0,T]\rightarrow {\mathbb {R}}$ be defined as $F(v):=\lim _n (1/v)\int _0^v(v-t)f_n(t)dt$. Note that we have $F(v)\ge 0~(v\in (0,T])$. Now the proof proceeds in several steps.

STEP1: We show $B(t)\ge 0~(a.e.t\in [0,T])$ and $B\in L^1([0,U];{\mathbb {R}})$. Since $\lim _n \int _v^U|f_n(t)-B(t)|dt=0~(v\in (0,U])$, by a standard diagonal argument, there exists a subsequence $\left\{ f_{n_k} \right\} _{k\ge 0}$ of $\left\{ f_n \right\} _{n\ge 0}$ such that $\lim _k f_{n_k}(t)=B(t)~(a.e.t\in [0,U])$. Since $f_n(t)\ge 0~(n\ge 0,~a.e.t\in [0,T])$, we obtain $B(t)\ge 0~(a.e.t\in [0,T])$. Moreover, in view of Fatou’s Lemma, for any $v\in (0,T]$, we have

$$\begin{aligned} {\mathbb {R}}\ni F(v)&=\lim _n \frac{1}{v}\int _0^v(v-t)f_n(t)dt=\lim _k \frac{1}{v}\int _0^v(v-t)f_{n_k}(t)dt\\&=\liminf _k \frac{1}{v}\int _0^v(v-t)f_{n_k}(t)dt \ge \frac{1}{v}\int _0^v(v-t)\liminf _k f_{n_k}(t)dt\\&=\frac{1}{v}\int _0^v(v-t)B(t)dt\ge \frac{1}{v}\int _0^{v/2}(v-t)B(t)dt\ge \frac{1}{2}\int _0^{v/2}B(t)dt, \end{aligned}$$

so $\int _0^{v/2}B(t)dt<+\infty ~(v\in (0,T])$. Keeping in mind $B\in L^1([v,U];{\mathbb {R}})~(v\in (0,U])$, we obtain $B\in L^1([0,U];{\mathbb {R}})$.

STEP2: We show that $F:(0,T]\rightarrow {\mathbb {R}_{\ge 0}}$ is non-decreasing and $F(+0)\in {\mathbb {R}_{\ge 0}}$ exists. Let $0<\varepsilon \le v\le T$ be arbitrary. We have

$$\begin{aligned} \frac{1}{v}\int _0^v(v-t)f_n(t)dt-\frac{1}{\varepsilon }\int _0^\varepsilon (\varepsilon -t)f_n(t)dt&=\int _\varepsilon ^vf_n+\frac{1}{\varepsilon }\int _0^\varepsilon tf_n-\frac{1}{v}\int _0^v tf_n\\&\ge \int _\varepsilon ^vf_n+\frac{1}{v}\int _0^\varepsilon tf_n-\frac{1}{v}\int _0^v tf_n\\&=\frac{1}{v}\int _\varepsilon ^v(v-t)f_n\ge 0. \end{aligned}$$

Taking $n\rightarrow +\infty$, we have $F(v)-F(\varepsilon )\ge 0$, i.e., $F(v)\ge F(\varepsilon )$. This implies that $F:(0,T]\rightarrow {\mathbb {R}_{\ge 0}}$ is non-decreasing. Then it follows that $F(+0)\in {\mathbb {R}_{\ge 0}}$ exists.

STEP3: In this step, we show $\lim _n \int _0^v f_n=C+\int _0^v B~(v\in (0,T])$ for some $C\in {\mathbb {R}}$. First, since $\lim _n \int _v^U|f_n(t)-B(t)|dt=0~(v\in (0,U])$, we have $\lim _n \int _u^v b(t)f_n(t)dt=\int _u^vb(t)B(t)dt$ for any $0<u\le v\le U$ and $b\in L^\infty ([u,v];{\mathbb {R}})$. Now let $0<\varepsilon <1$ and $0<v\le T$ be arbitrary. We have

$$\begin{aligned} \int _0^v(v-t)f_n(t)dt\rightarrow vF(v)~~and~~\int _0^{\varepsilon v}(\varepsilon v-t)f_n(t)dt\rightarrow \varepsilon vF(\varepsilon v)~~(n\rightarrow +\infty ). \end{aligned}$$

Since

$$\begin{aligned} \varepsilon \int _0^v(v-t)f_n(t)dt-\int _0^{\varepsilon v}(\varepsilon v-t)f_n(t)dt =\varepsilon v\int _{\varepsilon v}^v f_n-\int _{\varepsilon v}^v tf_n+(1-\varepsilon )\int _0^v tf_n, \end{aligned}$$

taking $n\rightarrow +\infty$, we obtain

$$\begin{aligned} \varepsilon vF(v)-\varepsilon vF(\varepsilon v)=\varepsilon v\int _{\varepsilon v}^v B-\int _{\varepsilon v}^v tB+(1-\varepsilon )\lim _n \int _0^v tf_n. \end{aligned}$$

(6)

This implies that $\lim _n \int _0^v tf_n(t)dt$ exists in ${\mathbb {R}}$ for any $v\in (0,T]$. Since (6) holds for any $0<\varepsilon <1$ and $0<v\le T$, we take $\varepsilon \downarrow 0$. Then

$$\begin{aligned} 0vF(v)-0vF(+0)=0v\int _0^v B-\int _0^v tB+\lim _n \int _0^v tf_n, \end{aligned}$$

so we obtain $\lim _n \int _0^v tf_n(t)dt=\int _0^v tB(t)dt~(v\in (0,T])$. Next, for $v\in (0,T]$, we have

$$\begin{aligned} \int _0^v f_n=\frac{1}{v}\int _0^v tf_n+\frac{1}{v}\int _0^v(v-t)f_n. \end{aligned}$$

Taking $n\rightarrow +\infty$, we have

$$\begin{aligned} \lim _n \int _0^v f_n=\frac{1}{v}\int _0^v tB+F(v)~(v\in (0,T]). \end{aligned}$$

(7)

This implies that $\lim _n \int _0^v f_n$ exists in ${\mathbb {R}}$ for any $v\in (0,T]$, and we have the equation (7). Next, let $0<v\le T$ be arbitrary. For $0<\varepsilon \le v$, we have $\int _0^vf_n=\int _0^\varepsilon f_n+\int _\varepsilon ^v f_n$. Taking $\lim _n$, and applying (7) to the term $\int _0^\varepsilon f_n$, we obtain

$$\begin{aligned} \lim _n \int _0^v f_n=\left( \frac{1}{\varepsilon }\int _0^\varepsilon tB+F(\varepsilon )\right) +\int _\varepsilon ^v B. \end{aligned}$$

We would like to take $\varepsilon \downarrow 0$. In view of $B(t)\ge 0~(a.e.t\in [0,T])$, we have

$$\begin{aligned} 0\le \frac{1}{\varepsilon }\int _0^\varepsilon tB(t)dt\le \frac{1}{\varepsilon }\int _0^\varepsilon \varepsilon B(t)dt=\int _0^\varepsilon B(t)dt\rightarrow 0~~(\varepsilon \downarrow 0), \end{aligned}$$

so we obtain $\lim _n \int _0^v f_n=C+\int _0^v B~(v\in (0,T])$ with $C:=F(+0)$.

STEP4: In this step, we complete the proof. Let $a\in I([0,U];{\mathbb {R}})$ and $v\in (0,U]$ be arbitrary. Let $\varepsilon >0$. There exists $0<\delta <\min \left\{ T,v \right\}$ such that $|a(t)-a(0)|<\varepsilon ~(0\le t<\delta )$. For any $0<\eta <\delta$, we have

$$\begin{aligned}&\int _0^v a(t)f_n(t)dt=\int _0^\eta a(t)f_n(t)dt+\int _\eta ^v a(t)f_n(t)dt, \\&\left| \int _0^\eta a(t)f_n(t)dt-\int _0^\eta a(0)f_n(t)dt\right| \le \varepsilon \int _0^\eta f_n(t)dt, \end{aligned}$$

so

$$\begin{aligned}&\int _0^v a(t)f_n(t)dt\le (\varepsilon +a(0))\int _0^\eta f_n(t)dt+\int _\eta ^v a(t)f_n(t)dt, \\&\int _0^v a(t)f_n(t)dt\ge (-\varepsilon +a(0))\int _0^\eta f_n(t)dt+\int _\eta ^v a(t)f_n(t)dt. \end{aligned}$$

Keeping in mind $0<\eta <T$, we have

$$\begin{aligned}&\limsup _n \int _0^v a(t)f_n(t)dt\le (\varepsilon +a(0))\left( \int _0^\eta B(t)dt+C\right) +\int _\eta ^v a(t)B(t)dt, \\&\liminf _n \int _0^v a(t)f_n(t)dt\ge (-\varepsilon +a(0))\left( \int _0^\eta B(t)dt+C\right) +\int _\eta ^v a(t)B(t)dt. \end{aligned}$$

Since $0<\eta <\delta$ is arbitrary, we can take $\eta \downarrow 0$, and we obtain

$$\begin{aligned}&\limsup _n \int _0^v a(t)f_n(t)dt\le (\varepsilon +a(0))C+\int _0^v a(t)B(t)dt, \\&\liminf _n \int _0^v a(t)f_n(t)dt\ge (-\varepsilon +a(0))C+\int _0^v a(t)B(t)dt. \end{aligned}$$

Since $\varepsilon >0$ is arbitrary, we obtain $\lim _n \int _0^v a(t)f_n(t)dt=a(0)C+\int _0^v a(t)B(t)dt$.

Thus we complete the proof, provided that $g(t)=0~(t\in [0,T])$. The general case easily follows from the above case by considering $\widetilde{f}_n,\widetilde{B}:[0,U]\rightarrow {\mathbb {R}}$ defined as $\widetilde{f}_n:=f_n-h$, $\widetilde{B}:=B-h$, and $h(t):=g(t)~(t\in [0,T]),~0~(t\in (T,U])$. $\square$

Lemma 5

Let $T>0$. Let $\left\{ f_n \right\} _{n\ge 0}\subset L^1_{loc}({\mathbb {R}_{\ge 0}};{\mathbb {R}})$. Let $B:{\mathbb {R}_{\ge 0}}\rightarrow {\mathbb {R}}$ satisfy $B\in L^1([u,v];{\mathbb {R}})~(0<u\le v)$. Assume that

• $\forall u,v\in {\mathbb {R}}_{>0}~\left[ ~ 0<u\le v~\Rightarrow ~ \lim _n \int _u^v|f_n(t)-B(t)|dt=0 ~\right]$,

• $\forall v\in (0,T]~\left[ ~ \lim _n (1/v)\int _0^v(v-t)f_n(t)dt~exists~in~{\mathbb {R}} ~\right]$,

• $\exists g\in L^1([0,T];{\mathbb {R}}),~\forall n\ge 0,~a.e.t\in [0,T]~\left[ ~ f_n(t)\ge g(t) ~\right]$.

Then we have $B\in L^1_{loc}({\mathbb {R}_{\ge 0}};{\mathbb {R}})$. Moreover, there exists $C\in {\mathbb {R}}$ such that

$$\begin{aligned} \forall a\in I({\mathbb {R}_{\ge 0}};{\mathbb {R}}),~\forall v\in {\mathbb {R}}_{>0}~\left[ ~ \lim _n \int _0^va(t)f_n(t)dt=\int _0^va(t)B(t)dt+Ca(0) ~\right] . \end{aligned}$$

Proof

This lemma easily follows from Lemma 4. $\square$

Remark 6

As we have mentioned in Remark 1, the Wiener–Ikehara theorem usually requires a constant $A\in {\mathbb {R}}$ to obtain $\lim _{x\rightarrow +\infty }f(x)=A$, while Theorem 2 does not require such A. In fact, the constant $C\in {\mathbb {R}}$ in Lemma 5 is the corresponding “A”.

4 Proof of Theorem 2

In this section, we prove Theorem 2.

Definition 2

We define $\mathrm{FK}(x):=(1-\cos x)/x^2=(\sin ^2(x/2))/(2(x/2)^2)~(x\ne 0),~1/2~(x=0)$ (Féjer kernel). Note that we have $\mathrm{FK}(-x)=\mathrm{FK}(x)\ge 0$ $(x\in {\mathbb {R}})$, $\mathrm{FK}\in L^1({\mathbb {R}};{\mathbb {R}_{\ge 0}})\cap L^\infty ({\mathbb {R}};{\mathbb {R}_{\ge 0}})$, $\int _{-\infty }^\infty \mathrm{FK}(x)dx>0$, and $\lim _{|x|\rightarrow +\infty }\mathrm{FK}(x)=0$.

Lemma 6

Let $f:{\mathbb {R}_{\ge 0}}\rightarrow {\mathbb {R}}$ be measurable such that the Lebesgue integral $F(s):=\int _0^\infty f(x)e^{-sx}dx$ converges in ${\mathbb {C}}$ for $\mathfrak {R}(s)>0$. Then F(s) is analytic on $\mathfrak {R}(s)>0$. Moreover, for any $\sigma ,v>0$ and $y\in {\mathbb {R}}$, we have

$$\begin{aligned}&\frac{2}{v}\int _0^v(v-t) \mathfrak {R}F(\sigma +it)\cos ty dt\\&\quad =\int _0^\infty f(x)e^{-\sigma x}v\mathrm{FK}(v(x-y))dx+\int _0^\infty f(x)e^{-\sigma x}v\mathrm{FK}(v(x+y))dx. \end{aligned}$$

Proof

The analyticity of F is well-known. Next, let $I(\sigma ,v,y):=\int _0^\infty f(x)e^{-\sigma x}v\mathrm{FK}(v(x-y))dx$, for short. Let $\sigma ,v>0$ and $y\in {\mathbb {R}}$. In view of Fubini’s theorem, we have

$$\begin{aligned} \int _0^v\int _{-u}^u F(\sigma +it)e^{ity}dtdu&=\int _0^v\int _{-u}^u \int _0^\infty f(x)e^{-\sigma x}e^{-it(x-y)}dxdtdu=\cdots \\&=\int _0^\infty f(x)e^{-\sigma x}2v^2\mathrm{FK}(v(x-y))dx=2vI(\sigma ,v,y). \end{aligned}$$

In short,

$$\begin{aligned} \frac{1}{2v}\int _0^v\int _{-u}^u F(\sigma +it)e^{ity}dtdu=I(\sigma ,v,y). \end{aligned}$$

Since $y\in {\mathbb {R}}$ is arbitrary, we can replace y with $-y$, and we obtain a corresponding equation. Adding them, and keeping in mind $e^{ity}+e^{-ity}=2\cos ty$, we have

$$\begin{aligned} \frac{1}{v}\int _0^v\int _{-u}^u F(\sigma +it)\cos tydtdu=I(\sigma ,v,y)+I(\sigma ,v,-y). \end{aligned}$$

Since the right hand side is in ${\mathbb {R}}$, so is the left hand side. This implies

$$\begin{aligned} \frac{1}{v}\int _0^v\int _{-u}^u \mathfrak {R}F(\sigma +it)\cos tydtdu=I(\sigma ,v,y)+I(\sigma ,v,-y). \end{aligned}$$

In general, we have $\int _{-u}^u \alpha (t)dt=2\int _0^u \alpha (t)dt$, provided that $\alpha (-t)=\alpha (t)$. In view of $\mathfrak {R}F(\sigma -it)\cos (-ty)=\mathfrak {R}F(\sigma +it)\cos ty$, we have

$$\begin{aligned} \frac{2}{v}\int _0^v\int _0^u \mathfrak {R}F(\sigma +it)\cos tydtdu=I(\sigma ,v,y)+I(\sigma ,v,-y). \end{aligned}$$

In general, we have $\int _0^v\int _0^u \alpha (t)dtdu=\int _0^v(v-t)\alpha (t)dt$, so

$$\begin{aligned} \frac{2}{v}\int _0^v(v-t)\mathfrak {R}F(\sigma +it)\cos ty dt=I(\sigma ,v,y)+I(\sigma ,v,-y). \end{aligned}$$

Thus we complete the proof. $\square$

Now we prove Theorem 2.

Proof

Let $f_n(t):=\mathfrak {R}F(\sigma _n+it)~(t\in {\mathbb {R}_{\ge 0}},~n\ge 0)$. Since $f_n(t)$ is continuous on $t\in {\mathbb {R}_{\ge 0}}$, we have $f_n\in L^1_{loc}({\mathbb {R}_{\ge 0}};{\mathbb {R}})$. Moreover, by (i) and (ii) of Theorem 2, we have

• (a) $\forall u,v\in {\mathbb {R}}_{>0}~\left[ ~ 0<u\le v~\Rightarrow ~ \lim _N\sup _{n,m\ge N} \int _u^v|f_n(t)-f_m(t)|dt=0 ~\right]$,

• (b) $\forall n\ge 0,~a.e.t\in [0,T]~\left[ ~ f_n(t)\ge g(t) ~\right]$.

The condition (a) states that $\left\{ f_n \right\} _{n\ge 0}$ is a cauchy sequence in $L^1([u,v];{\mathbb {R}})$ for any $0<u\le v$. Therefore, by a standard diagonal argument, there exists a subsequence $\left\{ f_{n_k} \right\} _{k\ge 0}$ of $\left\{ f_n \right\} _{n\ge 0}$ and $B:{\mathbb {R}_{\ge 0}}\rightarrow {\mathbb {R}}$ with $B\in L^1([u,v];{\mathbb {R}})~(0<u\le v)$ such that $\lim _k \int _u^v|f_{n_k}(t)-B(t)|dt=0~(0<u\le v)$. At this point, we have the following:

• $\left\{ f_{n_k} \right\} _{k\ge 0}\subset L^1_{loc}({\mathbb {R}_{\ge 0}};{\mathbb {R}})$, $g\in L^1([0,T];{\mathbb {R}})$,

• $B:{\mathbb {R}_{\ge 0}}\rightarrow {\mathbb {R}},~B\in L^1([u,v];{\mathbb {R}})~(0<u\le v)$,

• $\forall u,v\in {\mathbb {R}}_{>0}~\left[ ~ 0<u\le v~\Rightarrow ~ \lim _k \int _u^v|f_{n_k}(t)-B(t)|dt=0 ~\right]$,

• $\forall k\ge 0,~a.e.t\in [0,T]~\left[ ~ f_{n_k}(t)\ge g(t) ~\right]$.

Next, we apply Lemma 6 with $\sigma ,v>0$ and $y=0$. Then

$$\begin{aligned} \frac{2}{v}\int _0^v(v-t) \mathfrak {R}F(\sigma +it)dt=2\int _0^\infty f(x)e^{-\sigma x}v\mathrm{FK}(vx)dx. \end{aligned}$$

Taking $\sigma =\sigma _{n_k}$, we obtain

$$\begin{aligned} \frac{1}{v}\int _0^v (v-t)f_{n_k}(t)dt=\int _0^\infty f(x)e^{-\sigma _{n_k} x}v\mathrm{FK}(vx)dx~~(k\ge 0,~v>0). \end{aligned}$$

We would like to take $k\rightarrow +\infty$. Since $\int _0^\infty |f(x)|/(x+1)^2dx<+\infty$ and $\mathrm{FK}(x)=(1-\cos x)/x^2~(x\ne 0)$, we can show $\int _0^\infty |f(x)|v\mathrm{FK}(vx)dx<+\infty$. By the DCT, we have

$$\begin{aligned} \lim _k \frac{1}{v}\int _0^v (v-t)f_{n_k}(t)dt=\int _0^\infty f(x)v\mathrm{FK}(vx)dx\in {\mathbb {R}}~~(v>0). \end{aligned}$$

In particular, we have

• $\forall v\in (0,T]~\left[ ~ \lim _k (1/v)\int _0^v(v-t)f_{n_k}(t)dt~exists~in~{\mathbb {R}} ~\right]$.

Then we can apply Lemma 5, so $B\in L^1_{loc}({\mathbb {R}_{\ge 0}};{\mathbb {R}})$ (this is important). Moreover, the same lemma implies that there exists $C\in {\mathbb {R}}$ such that

$$\begin{aligned} \forall a\in I({\mathbb {R}_{\ge 0}};{\mathbb {R}}),~\forall v\in {\mathbb {R}}_{>0}~\left[ ~ \lim _k \int _0^va(t)f_{n_k}(t)dt=\int _0^va(t)B(t)dt+Ca(0) ~\right] . \end{aligned}$$

(8)

Now let $k\ge 0$, $v>0$, and $y\in {\mathbb {R}}$. By Lemma 6 with $\sigma =\sigma _{n_k}$, we have

$$\begin{aligned}&\frac{2}{v}\int _0^v (v-t)f_{n_k}(t)\cos ty dt \nonumber \\&\quad =\int _0^\infty f(x)e^{-\sigma _{n_k}x}v\mathrm{FK}(v(x-y))dx+\int _0^\infty f(x)e^{-\sigma _{n_k}x}v\mathrm{FK}(v(x+y))dx. \end{aligned}$$

(9)

We would like to take $k\rightarrow +\infty$. Keeping in mind (8), we have

$$\begin{aligned}&\frac{2}{v}\int _0^v (v-t)f_{n_k}(t)\cos ty dt =2\int _0^v (\cos ty)f_{n_k}(t)dt-\frac{2}{v}\int _0^v (t\cos ty)f_{n_k}(t) dt\\&\quad \rightarrow 2\left( \int _0^v(\cos ty)B(t)dt+C\cos 0y\right) -\frac{2}{v}\left( \int _0^v(t\cos ty)B(t)dt+C0\cos 0y\right) \\&\quad =\frac{2}{v}\int _0^v(v-t)B(t)\cos tydt+2C~~(k\rightarrow +\infty ). \end{aligned}$$

Moreover, by the DCT, the right hand side of (9) converges to $\int _0^\infty f(x)v\mathrm{FK}(v(x-y))dx+\int _0^\infty f(x)v\mathrm{FK}(v(x+y))dx$. Then we obtain

$$\begin{aligned}&\frac{2}{v}\int _0^v(v-t)B(t)\cos tydt+2C\\&\quad =\int _0^\infty f(x)v\mathrm{FK}(v(x-y))dx+\int _0^\infty f(x)v\mathrm{FK}(v(x+y))dx~~(v>0,~y\in {\mathbb {R}}). \end{aligned}$$

For each $v>0$, we would like to take $y\rightarrow +\infty$. Since $\mathrm{FK}(x)=(1-\cos x)/x^2~(x\ne 0)$, for any $v>0$ and $y\ge 1$ we have

$$\begin{aligned} \left| \int _0^\infty f(x)v\mathrm{FK}(v(x+y))dx\right| \le \int _0^\infty |f(x)|v\frac{2}{(v(x+y))^2}dx=\frac{2}{v}\int _0^\infty \frac{|f(x)|}{(x+y)^2}dx. \end{aligned}$$

Since $\int _0^\infty |f(x)|/(x+1)^2dx<+\infty$, it follows from the DCT that $\lim _{y\rightarrow +\infty }\int _0^\infty |f(x)|/(x+y)^2dx=0$, so $\lim _{y\rightarrow +\infty }\int _0^\infty f(x)v\mathrm{FK}(v(x+y))dx=0$. Moreover, since $B\in L^1_{loc}({\mathbb {R}_{\ge 0}};{\mathbb {R}})$, we have $\lim _{y\rightarrow +\infty }(2/v)\int _0^v(v-t)B(t)\cos tydt=0$ by Riemann-Lebesgue Lemma. Hence

$$\begin{aligned} \lim _{y\rightarrow +\infty }\int _0^\infty f(x)v\mathrm{FK}(v(x-y))dx=2C~(\forall v>0). \end{aligned}$$

By Corollary 1, we conclude that $\lim _{x\rightarrow +\infty }f(x)$ exists in ${\mathbb {R}}$. $\square$

5 Proof of Theorem 3

In this section, we prove Theorem 3.

Lemma 7

Let $a:{\mathbb {Z}_{\ge 0}}\rightarrow {\mathbb {R}}$ be a sequence such that $F(z):=\sum _{n=0}^\infty a_nz^n$ converges in ${\mathbb {C}}$ for any $|z|<1$. Then F(z) is analytic on $|z|<1$. Moreover, for any $r\in (0,1)$, $v>0$, and $m\in {\mathbb {Z}}$, we have

$$\begin{aligned}&\frac{2}{v}\int _0^v(v-t) \mathfrak {R}F(re^{it})\cos mt dt =\sum _{k=0}^\infty a_kr^k v\mathrm{FK}(v(k-m))\\&\quad +\sum _{k=0}^\infty a_kr^k v\mathrm{FK}(v(k+m)). \end{aligned}$$

Proof

The analyticity of F(z) is well-known. Next, let $I(r,v,m):=\sum _{k=0}^\infty a_kr^k v\mathrm{FK}(v(k-m))$, for short. Let $r\in (0,1)$, $v>0$, and $m\in {\mathbb {Z}}$. It is easy to verify $\int _0^v\int _{-u}^u e^{ikt}dtdu=2v^2\mathrm{FK}(vk)~(k\in {\mathbb {Z}})$. Then

$$\begin{aligned}&\int _0^v\int _{-u}^u F(re^{it})e^{-imt}dtdu=\sum _{k=0}^\infty a_kr^k \int _0^v\int _{-u}^u e^{i(k-m)t}dtdu\\&\quad =\sum _{k=0}^\infty a_kr^k 2v^2\mathrm{FK}(v(k-m))=2vI(r,v,m). \end{aligned}$$

In short,

$$\begin{aligned} \frac{1}{2v}\int _0^v\int _{-u}^u F(re^{it})e^{-imt}dtdu=I(r,v,m). \end{aligned}$$

Since $m\in {\mathbb {Z}}$ is arbitrary, we can replace m with $-m$, and we obtain a corresponding equation. Adding them, and keeping in mind $e^{itm}+e^{-itm}=2\cos mt$, we have

$$\begin{aligned} \frac{1}{v}\int _0^v\int _{-u}^u F(re^{it})\cos mtdtdu=I(r,v,m)+I(r,v,-m). \end{aligned}$$

Then, by an argument similar to that of Lemma 6, we complete the proof. $\square$

Lemma 8

Let $a:{\mathbb {Z}_{\ge 0}}\rightarrow {\mathbb {R}}$ satisfy the conditions of Lemma 7. Then for any $r\in (0,1)$ and $m\in {\mathbb {Z}}_{\ge 1}$, we have

$$\begin{aligned} a_mr^m=\frac{2}{\pi }\int _0^\pi \mathfrak {R}F(re^{it})\cos mt dt. \end{aligned}$$

Proof

Let $r\in (0,1)$ and $m\in {\mathbb {Z}}$. We have

$$\begin{aligned} \int _{-\pi }^\pi F(re^{it})e^{-imt}dt =\sum _{k=0}^\infty a_kr^k\int _{-\pi }^\pi e^{i(k-m)t}dt=0~(m<0),~2\pi a_mr^m~(m\ge 0). \end{aligned}$$

Then, for $m\ge 1$, we have $\int _{-\pi }^\pi F(re^{it})e^{-imt}dt=2\pi a_mr^m$ and $\int _{-\pi }^\pi F(re^{it})e^{imt}dt=0$. Adding them, we have $2\int _{-\pi }^\pi F(re^{it})\cos mt dt=2\pi a_mr^m$. Since the right hand side is in ${\mathbb {R}}$, we have $\int _{-\pi }^\pi \mathfrak {R}F(re^{it})\cos mt dt=\pi a_mr^m$. Then we have $2\int _0^\pi \mathfrak {R}F(re^{it})\cos mt dt=\pi a_mr^m$ by $\mathfrak {R}F(re^{-it})\cos m(-t)=\mathfrak {R}F(re^{it})\cos mt$. Thus we complete the proof. $\square$

Now we prove Theorem 3.

Proof

Let $f_n(t):=\mathfrak {R}F(r_ne^{it})~(t\in [0,\pi ],~n\ge 0)$. Since $f_n(t)$ is continuous on $t\in [0,\pi ]$, we have $f_n\in L^1([0,\pi ];{\mathbb {R}})$. Next, (i) of Theorem 3 implies $\lim _N\sup _{n,m\ge N} \int _v^\pi |f_n(t)-f_m(t)|dt=0~(v\in (0,\pi ])$. Therefore, by a standard diagonal argument, there exists a subsequence $\left\{ f_{n_k} \right\} _{k\ge 0}$ of $\left\{ f_n \right\} _{n\ge 0}$ and $B:[0,\pi ]\rightarrow {\mathbb {R}}$ with $B\in L^1([v,\pi ];{\mathbb {R}})~(v\in (0,\pi ])$ such that $\lim _k \int _v^\pi |f_{n_k}(t)-B(t)|dt=0~(v\in (0,\pi ])$. Now we make use of Lemma 7 and Lemma 4. By an argument similar to the corresponding proof of Theorem 2, we have $B\in L^1([0,\pi ];{\mathbb {R}})$, and there exists $C\in {\mathbb {R}}$ such that

$$\begin{aligned} \forall a\in I([0,\pi ];{\mathbb {R}}),~\forall v\in (0,\pi ]~\left[ ~ \lim _k \int _0^va(t)f_{n_k}(t)dt=\int _0^va(t)B(t)dt+Ca(0) ~\right] . \end{aligned}$$

(10)

Now let $k\ge 0$ and $m\in {\mathbb {Z}}_{\ge 1}$ be arbitrary. By Lemma 8 with $r=r_{n_k}$, we have $a_mr_{n_k}^m=(2/\pi )\int _0^\pi f_{n_k}(t)\cos mtdt$. We would like to take $k\rightarrow +\infty$. In view of (10), we have

$$\begin{aligned} a_m=\frac{2}{\pi }\left( \int _0^\pi B(t)\cos mtdt+C\cos m0\right) =\frac{2}{\pi }\int _0^\pi B(t)\cos mtdt+\frac{2C}{\pi }~~(m\in {\mathbb {Z}}_{\ge 1}). \end{aligned}$$

Since $B\in L^1([0,\pi ];{\mathbb {R}})$, we have $\lim _{m\rightarrow +\infty }\int _0^\pi B(t)\cos mtdt=0$ by Riemann-Lebesgue Lemma. Consequently, we obtain $\lim _m a_m=2C/\pi$. $\square$

6 Application 1

In this section, we provide an application of Theorem 3. Our starting point is the following theorem:

Theorem 4

Let $\left\{ q_n \right\} _{n\ge 0}\subset [0,1]$ satisfy $q_0=0,~\sum _{n=0}^\infty q_n=1$, and

$\gcd \left\{ n\ge 1\mid q_n\ne 0 \right\} =1$. Let $\left\{ p_n \right\} _{n\ge 0}\subset {\mathbb {R}}$ be defined as $p_0:=1$ and $p_n:=\sum _{k=0}^n q_kp_{n-k}=\sum _{k=1}^n q_kp_{n-k}~(n\ge 1)$. Then we have $\lim _n p_n=1/\sum _{k=0}^\infty kq_k$, where we define $1/\sum _{k=0}^\infty kq_k=0$ if $\sum _{k=0}^\infty kq_k=+\infty$.

This theorem was first proved by [7], and was called Basic Limit Theorem in renewal theory (e.g., [8]). A basic generalization of this theorem is Blackwell’s renewal theorem (BRT, for short) [1, 2]. In this direction, Theorem 4 can be understood as a discrete case of BRT.

Many proofs of BRT are known. Some references can be found in [6], where [6] also provides a proof of BRT. In this section, we also give a proof of BRT (of discrete case), i.e., we prove Theorem 4.

Proof

For $a:{\mathbb {Z}_{\ge 0}}\rightarrow {\mathbb {C}}$ and $z\in {\mathbb {C}}$, we define $F_a(z):=\sum _{n=0}^\infty a_nz^n$, provided that the right hand side converges in ${\mathbb {C}}$. Next, we define $B:=\left\{ z\in {\mathbb {C}}\mid |z|<1 \right\}$ and $D:=\left\{ z\in {\mathbb {C}}\mid |z|\le 1 \right\}$. Now let p and q satisfy the conditions of Theorem 4. We would like to show $\lim _n p_n=1/\sum _{k=0}^\infty kq_k$. First, since $p_0=1$ and $p_n=\sum _{k=1}^n q_kp_{n-k}~(n\ge 1)$, we can easily show $0\le p_n\le 1$ by induction on $n\ge 0$. Then $F_p(z)$ is well-defined for $z\in B$. Moreover, since $\sum _n |q_n|=\sum _n q_n=1<+\infty$, $F_q(z)$ is well-defined for $z\in D$. Note that $F_q(z)$ is analytic on $z\in B$ and continuous on $z\in D$. Next, we can show $F_q(z)\ne 1~(z\in D-\left\{ 1 \right\} )$ with a little effort by using the condition $\gcd \left\{ n\ge 1\mid q_n\ne 0 \right\} =1$. Moreover, since $p_0=1$ and $p_n=\sum _{k=0}^n q_kp_{n-k}~(n\ge 1)$, we can easily show that $F_p(z)=1+F_p(z)F_q(z)~(z\in B)$, so $F_p(z)=(1-F_q(z))^{-1}~(z\in B)$. Let $r_n:=1-1/(n+2)~(n\ge 0)$ and $T:=1$. We would like to show

• (a) $\forall v\in (0,\pi ]~\left[ ~ \lim _N\sup _{n,m\ge N}\int _v^\pi |\mathfrak {R}F_p(r_ne^{it})-\mathfrak {R}F_p(r_me^{it})|dt=0 ~\right]$,

• (b) $\forall n\ge 0,~a.e.t\in [0,T]~\left[ ~ \mathfrak {R}F_p(r_ne^{it})\ge 0 ~\right]$.

We first show (a). Since $F_p(z)=(1-F_q(z))^{-1}~(z\in B)$, it suffices to show that

$$\begin{aligned} \lim _N\sup _{n,m\ge N}\int _v^\pi |\mathfrak {R}((1-F_q(r_ne^{it}))^{-1})-\mathfrak {R}((1-F_q(r_me^{it}))^{-1})|dt=0~(v\in (0,\pi ]). \end{aligned}$$

(11)

Since $F_q(z)\ne 1~(z\in D-\left\{ 1 \right\} )$, we can define $(1-F_q(z))^{-1}$ on $z\in D-\left\{ 1 \right\}$. Since $F_q(z)$ is continuous on D, it follows that $(1-F_q(z))^{-1}$ is continuous on $D-\left\{ 1 \right\}$. In particular, $(1-F_q(z))^{-1}$ is uniformly continuous on any compact $K\subset {\mathbb {C}}$ with $K\subset D-\left\{ 1 \right\}$. Then we can easily verify (11). Thus we obtain (a). Next, let $r\in (0,1)$ and $t\in (0,T]$. Let $x:=1-\sum _{n=0}^\infty q_nr^n\cos nt$ and $y:=\sum _{n=0}^\infty q_nr^n\sin nt$. Then $F_p(re^{it})=(1-F_q(re^{it}))^{-1}=(x-iy)^{-1}$, so $\mathfrak {R}F_p(re^{it})=x/(x^2+y^2)$. Since $x=1-\sum _{n=0}^\infty q_nr^n\cos nt\ge 1-\sum _{n=0}^\infty q_n=0$, we have $x\ge 0$, so $\mathfrak {R}F_p(re^{it})\ge 0$. Then we trivially obtain (b). Moreover, since $0\le p_n\le 1$, we have $\sum _{n=0}^\infty |p_n|/(n+1)^2<+\infty$. Thus we can apply Theorem 3, and $A:=\lim _n p_n\in {\mathbb {R}}$ exists. Since $F_p(r)=\sum _{n=0}^\infty p_nr^n~(0<r<1)$, we can easily show $\lim _{r\uparrow 1}(1-r)F_p(r)=A$. Next, we show $\lim _{r\uparrow 1}(1-r)F_p(r)=1/\sum _{k=0}^\infty kq_k$. Let $r\in (0,1)$. Since $1-F_q(r)=1-\sum _{n=0}^\infty q_nr^n=\sum _{n=0}^\infty q_n-\sum _{n=0}^\infty q_nr^n =\sum _{n=1}^\infty q_n(1-r^n)$, we have $(1-F_q(r))/(1-r)=\sum _{n=1}^\infty q_n(1+r+\cdots +r^{n-1})$. For any $n\ge 1$, we have $0\le q_n(1+r+\cdots +r^{n-1})\uparrow nq_n~(r\uparrow 1)$. By the monotone convergence theorem (for series), we have $\lim _{r\uparrow 1}(1-F_q(r))/(1-r)=\sum _{k=1}^\infty kq_k~(\in (0,+\infty ])$. In view of $F_p(r)=1/(1-F_q(r))~(0<r<1)$, we have $\lim _{r\uparrow 1}(1-r)F_p(r)=1/\sum _{k=1}^\infty kq_k$. Thus we obtain $A=1/\sum _{k=1}^\infty kq_k$, and we complete the proof. $\square$

7 Application 2

In this section, we provide an application of Theorem 2. Specifically, we provide a Dirichlet series version of Theorem 4. Since the sequence p of Theorem 4 satisfies $p_n=\sum _{k=0}^n q_kp_{n-k}~(n\ge 1)$ with $q_0=0$, and since $\sum _{k=0}^n q_kp_{n-k}$ is a normal convolution of p and q, we simply replace it with Dirichlet convolution. Therefore, the definition of p should be $\left\{ p_n \right\} _{n\ge 1}\subset {\mathbb {R}},~p_1:=1,~p_n:=\sum _{d|n}q_dp_{n/d}~(n\ge 2)$ with $q_1=0$. The corresponding conditions of q should be at least $q\ge 0$, $q_1=0$, and $\sum _{n=1}^\infty q_n/n^\alpha =1$ for some $\alpha \in {\mathbb {R}}$. In this direction, we obtain the following theorem:

Theorem 5

(A Dirichlet type of Theorem 4) Let $q:{\mathbb {Z}}_{\ge 1}\rightarrow {\mathbb {R}_{\ge 0}}$ satisfy

$$\begin{aligned} q_1=0,~\sum _{n=1}^\infty \frac{q_n}{n}=1,~\forall d\in {\mathbb {Z}}_{\ge 2}~\left[ ~ \left\{ n\ge 1\mid q_n\ne 0 \right\} -\left\{ d^k\mid k\ge 1 \right\} \ne \emptyset ~\right] . \end{aligned}$$

(12)

Let $p:{\mathbb {Z}}_{\ge 1}\rightarrow {\mathbb {R}}$ be defined as

$$\begin{aligned} p_1:=1,~p_n:=\sum _{d|n}q_dp_{n/d}=\sum _{2\le d|n}q_dp_{n/d}~(n\ge 2). \end{aligned}$$

(13)

Then we have $\lim _n (1/n)\sum _{k=1}^n p_k=1/\sum _{n=1}^\infty (q_n/n)\log n$, where we define

$1/\sum _{n=1}^\infty (q_n/n)\log n=0$ if $\sum _{n=1}^\infty (q_n/n)\log n=+\infty$.

Remark 7

In this theorem, the direct limit “$\lim _n p_n=1/\sum _{n=1}^\infty (q_n/n)\log n$” does not hold in general. For example, let q satisfy $q_n=0$ for any prime number n. Then, for any prime number n, we have $p_n=q_1p_n+q_np_1=0p_n+0p_1=0$. Therefore, if $\lim _n p_n$ exists, then it should be $\lim _n p_n=0$, so if in addition $\sum _{n=1}^\infty (q_n/n)\log n<+\infty$, then we can not obtain $\lim _n p_n=1/\sum _{n=1}^\infty (q_n/n)\log n$. In this direction, we can actually construct an example of p and q which satisfy the conditions of Theorem 5 but $\lim _n p_n=1/\sum _{n=1}^\infty (q_n/n)\log n$ does not hold.

To prove Theorem 5, we need some lemmas.

Lemma 9

Let $a,b\in {\mathbb {Z}}_{\ge 2}$ and $x,y\in {\mathbb {Z}}_{\ge 1}$ satisfy $a^x=b^y$. Then there exists $d\in {\mathbb {Z}}_{\ge 2}$ such that $a,b\in \left\{ d^k\mid k\ge 1 \right\}$.

Proof

We can write $x=gx_1$, $y=gy_1$, $x_1,y_1\in {\mathbb {Z}}_{\ge 1}$, $g=\gcd (x,y)$, and $\gcd (x_1,y_1)=1$. Since $a^x=b^y$, we have $a^{gx_1}=b^{gy_1}$, so $a^{x_1}=b^{y_1}$. By the equality $a^{x_1}=b^{y_1}$, we have $\left\{ p\mid p~is~a~prime~with~p|a \right\} =\left\{ p\mid p~is~a~prime~with~p|b \right\}$. Then we can write $a=p_1^{e_1}\cdots p_k^{e_k}$ and $b=p_1^{f_1}\cdots p_k^{f_k}$ for some $k\ge 1$ and some primes $p_1,\ldots ,p_k$ and some $e_1,\ldots ,e_k,f_1,\ldots ,f_k\in {\mathbb {Z}}_{\ge 1}$. Since $a^{x_1}=b^{y_1}$, we have $\prod _i p_i^{e_ix_1}=\prod _i p_i^{f_iy_1}$, so $e_ix_1=f_iy_1$ for each $i\in [1,k]$. Since $\gcd (x_1,y_1)=1$, we have $x_1|f_i$, so we can write $f_i=x_1l_i$ for some $l_i\ge 1$. Then $e_i=y_1l_i$. Thus $a=\prod _i p_i^{e_i}=\prod _i p_i^{y_1l_i}=(\prod _i p_i^{l_i})^{y_1}$ and $b=\prod _i p_i^{f_i}=\prod _i p_i^{x_1l_i}=(\prod _i p_i^{l_i})^{x_1}$. Hence, $d:=\prod _i p_i^{l_i}\in {\mathbb {Z}}_{\ge 2}$ satisfies the claim. $\square$

Lemma 10

For $n,m\in {\mathbb {Z}}_{\ge 2}$, we define $n\sim m$ if and only if $n^x=m^y$ for some $x,y\ge 1$. It is easy to show that $\sim$ defines an equivalence relation on ${\mathbb {Z}}_{\ge 2}$. For $n\in {\mathbb {Z}}_{\ge 2}$, we define $C(n):=\left\{ m\in {\mathbb {Z}}_{\ge 2}\mid n\sim m \right\}$. Then we have the following:

$$\begin{aligned} \forall n\in {\mathbb {Z}}_{\ge 2},~\exists d\in {\mathbb {Z}}_{\ge 2}~\left[ ~ C(n)\subset \left\{ d^k\mid k\ge 1 \right\} ~\right] . \end{aligned}$$

Proof

In view of Lemma 9, we trivially obtain the following:

$$\begin{aligned} \forall n,m\in {\mathbb {Z}}_{\ge 2}~\left[ ~ n\sim m~\Rightarrow ~ \exists g\in {\mathbb {Z}}_{\ge 2}~\left[ ~ n,m\in \left\{ g^k\mid k\ge 1 \right\} ~\right] ~\right] . \end{aligned}$$

(14)

Now let $n\ge 2$ be arbitrary. Since $n\in C(n)$, we have $C(n)\ne \emptyset$. Let $d:=\min C(n)$. Then $d\in C(n)$, so $d\sim n$. Now we show $C(n)\subset \left\{ d^k\mid k\ge 1 \right\}$. Let $m\in C(n)$ be arbitrary. Then $n\sim m$. Since $d\sim n$, we have $d\sim m$. In view of (14), we have $d,m\in \left\{ g^k\mid k\ge 1 \right\}$ for some $g\ge 2$. Then $d=g^l$ for some $l\ge 1$. Since $g\sim g^l$, we have $g\sim d$. Since $d\sim n$, we have $g\sim n$, so $g\in C(n)$. Since $d=\min C(n)$, we have $d\le g$. Since $d=g^l\ge g$, we obtain $g\le d\le g$, so $g=d$. Then $\left\{ g^k\mid k\ge 1 \right\} =\left\{ d^k\mid k\ge 1 \right\}$. Keeping in mind $m\in \left\{ g^k\mid k\ge 1 \right\}$, we obtain $m\in \left\{ d^k\mid k\ge 1 \right\}$. Hence $C(n)\subset \left\{ d^k\mid k\ge 1 \right\}$. $\square$

Lemma 11

Let q satisfy (12). Then we have $\sum _{n=1}^\infty (q_n/n^\sigma )\cos (t\log n)<1$ for any $(\sigma ,t)\in {\mathbb {R}}_{\ge 1}\times {\mathbb {R}}-\left\{ (1,0) \right\}$.

Proof

Let $A:=\left\{ n\ge 1\mid q_n\ne 0 \right\} =\left\{ n\ge 2\mid q_n\ne 0 \right\} \subset {\mathbb {Z}}_{\ge 2}$. Since $\sum _{n=1}^\infty (q_n/n)=1$, we have $A\ne \emptyset$. Next, let $(\sigma ,t)\in {\mathbb {R}}_{\ge 1}\times {\mathbb {R}}-\left\{ (1,0) \right\}$. We show $\sum _{n=1}^\infty (q_n/n^\sigma )\cos (t\log n)<1$. If $\sigma >1$, then we trivially obtain it. Now we may assume $\sigma =1$. Since $(\sigma ,t)\ne (1,0)$, we have $t\ne 0$. Suppose $\sum _{n=1}^\infty (q_n/n)\cos (t\log n)\ge 1$ for the sake of contradiction. Since $\cos ((-t)\log n)=\cos (t\log n)$, we may assume $t>0$. In view of $1\le \sum _{n=1}^\infty (q_n/n)\cos (t\log n)\le \sum _{n=1}^\infty (q_n/n)=1$, we must have $(q_n/n)\cos (t\log n)=(q_n/n)~(n\ge 1)$, so $[q_n=0~or~t\log n\in 2\pi {\mathbb {Z}}]$ for each $n\ge 1$. Since $q_n\ne 0~(n\in A)$, we obtain $t\log n\in 2\pi {\mathbb {Z}}~(n\in A)$. Then there exists $k_n\in {\mathbb {Z}}$ such that $t\log n=2\pi k_n$ for $n\in A$. Since $t>0$ and $A\subset {\mathbb {Z}}_{\ge 2}$, we have $k_n\ge 1~(n\in A)$. Let $C:=t/2\pi$. Then we have $k_n/\log n=C~(n\in A)$. Therefore, for any $n,m\in A$, we have $k_n/\log n=C=k_m/\log m$, so $k_n\log m=k_m\log n$, i.e., $m^{k_n}=n^{k_m}$. Since $n,m\in A\subset {\mathbb {Z}}_{\ge 2}$ and $k_n,k_m\ge 1$, we have $n\sim m$. This implies

$$\begin{aligned} \forall n,m\in A~\left[ ~ n\sim m ~\right] . \end{aligned}$$

(15)

Since $A\ne \emptyset$, we can take $n_1\in A$. By (15), we have $A\subset C(n_1)$. By Lemma 10, we have $C(n_1)\subset \left\{ d^k\mid k\ge 1 \right\}$ for some $d\ge 2$. Then $A\subset \left\{ d^k\mid k\ge 1 \right\}$, i.e., $\left\{ n\ge 1\mid q_n\ne 0 \right\} \subset \left\{ d^k\mid k\ge 1 \right\}$. This contradicts (12). Thus we complete the proof. $\square$

Definition 3

For $\sigma \in {\mathbb {R}}$, we define ${\mathbb {C}}_{\ge \sigma }:=\left\{ s\in {\mathbb {C}}\mid \mathfrak {R}(s)\ge \sigma \right\}$. We also define ${\mathbb {C}}_{>\sigma }$ in a similar manner. Next, for $a:{\mathbb {Z}}_{\ge 1}\rightarrow {\mathbb {C}}$ and $s\in {\mathbb {C}}$, we define $\zeta _a(s):=\sum _{n=1}^\infty a_n/n^s$ if the right hand side converges in ${\mathbb {C}}$. For $a:{\mathbb {Z}}_{\ge 1}\rightarrow {\mathbb {C}}$, we define $S_a(x):=\sum _{n\le x}a_n~(x\in {\mathbb {R}}_{\ge 1})$. For $a,b:{\mathbb {Z}}_{\ge 1}\rightarrow {\mathbb {C}}$, we define $(a*b)_n:=\sum _{d|n}a_db_{n/d}~(n\ge 1)$. Next, we define $I:{\mathbb {Z}}_{\ge 1}\rightarrow {\mathbb {R}}$ as $I_1:=1,~I_n:=0~(n\ge 2)$.

Lemma 12

Let p and q satisfy the conditions of Theorem 5. Then $p\ge 0$ and $S_p(x)\le 2x~(x\in {\mathbb {R}}_{\ge 1})$.

Proof

Let $I:{\mathbb {Z}}_{\ge 1}\rightarrow {\mathbb {R}}$ be defined as Definition 3. We can easily verify $p\ge 0$ and $p=I+p*q$. Next, for any $a,b:{\mathbb {Z}}_{\ge 1}\rightarrow {\mathbb {C}}$ and $x\in {\mathbb {R}}_{\ge 1}$, we can show $S_{a*b}(x)=\sum _{n\le x}a_nS_b(x/n)=\sum _{n\le x}b_nS_a(x/n)$. Then it follows from $p=I+p*q$ that $S_p(x)=S_I(x)+S_{p*q}(x)=1+\sum _{n\le x}q_nS_p(x/n)~(x\in {\mathbb {R}}_{\ge 1})$. Let $f(x):=S_p(x)/x$. Then

$$\begin{aligned} f(x)=\frac{1}{x}+\sum _{n\le x}\frac{q_n}{n}f\left( \frac{x}{n}\right) ~(x\in {\mathbb {R}}_{\ge 1}). \end{aligned}$$

(16)

We have only to show $f(x)\le 2~(x\in {\mathbb {R}}_{\ge 1})$. It suffices to show

$$\begin{aligned} \forall k\ge 1,~\forall x\in [1,2^k)~\left[ ~ f(x)\le 2-2^{-(k-1)} ~\right] . \end{aligned}$$

(17)

The proof is by induction on $k\ge 1$. The case $k=1$ is trivial in view of $S_p(x)=p_1=1~(1\le x<2)$. Next, let $k\ge 1$ satisfy (17). Let $x\in [1,2^{k+1})$ be arbitrary. If $x\in [1,2^k)$, then $f(x)\le 2-2^{-(k-1)}\le 2-2^{-k}$. Now we may assume $x\in [2^k,2^{k+1})$. We have $\max _{2\le n\le x}f(x/n)=f(x/m)$ for some $2\le m\le x$. Then in view of (16) and $q_1=0$, we have $f(x)=1/x+\sum _{2\le n\le x}(q_n/n)f(x/n) \le 1/x+f(x/m)\sum _{2\le n\le x}(q_n/n)\le 1/x+f(x/m)$, i.e., $f(x)\le 1/x+f(x/m)$. Since $1\le x/m\le x/2<2^k$, we have $f(x/m)\le 2-2^{-(k-1)}$ by assumption on k. Since $x\in [2^k,2^{k+1})$, we have $1/x\le 2^{-k}$, so $f(x)\le 1/x+f(x/m)\le 2^{-k}+2-2^{-(k-1)}=2-2^{-k}$. Thus we obtain (17) by induction. $\square$

Now we prove Theorem 5.

Proof

Let p and q satisfy the conditions of Theorem 5. We would like to show $\lim _n (1/n)\sum _{k=1}^n p_n=1/\sum _{n=1}^\infty (q_n/n)\log n$. Since $q\ge 0$ and $\sum _{n=1}^\infty q_n/n=1<+\infty$, the series $\sum _{n=1}^\infty q_n/n^s$ is absolutely convergent (AC, for short) for $s\in {\mathbb {C}}_{\ge 1}$, so $\zeta _q(s)$ is well-defined for $s\in {\mathbb {C}}_{\ge 1}$. Note that $\zeta _q(s)$ is analytic on $s\in {\mathbb {C}}_{>1}$ and continuous on $s\in {\mathbb {C}}_{\ge 1}$. Moreover, by Lemma 11, we have $\zeta _q(s)\ne 1~(s\in {\mathbb {C}}_{\ge 1}-\left\{ 1 \right\} )$. Next, since $0\le S_p(x)/x\le 2$, we have $\sum _{n\le x}p_n/n^s=S_p(x)/x^s+s\int _1^xS_p(y)/y^{s+1}dy \rightarrow 0+s\int _1^\infty S_p(y)/y^{s+1}dy~(x\rightarrow +\infty )$ for any $s\in {\mathbb {C}}_{>1}$. This implies that $\zeta _p(s)$ is well-defined for $s\in {\mathbb {C}}_{>1}$ with $\zeta _p(s)=s\int _1^\infty S_p(x)/x^{s+1}dx$. Then, for $\sigma >1$ and $t\in {\mathbb {R}}$, we have $\sum _{n=1}^\infty |p_n/n^{\sigma +it}|=\sum _{n=1}^\infty p_n/n^\sigma =\zeta _p(\sigma )\in {\mathbb {R}}$, so $\sum _{n=1}^\infty p_n/n^s$ is AC for $s\in {\mathbb {C}}_{>1}$. Since $\sum _{n=1}^\infty q_n/n^s$ is AC for $s\in {\mathbb {C}}_{\ge 1}$, we can show that $\sum _{n=1}^\infty (p*q)_n/n^s$ is AC for $s\in {\mathbb {C}}_{>1}$ with $\sum _{n=1}^\infty (p*q)_n/n^s=(\sum _{n=1}^\infty p_n/n^s)(\sum _{n=1}^\infty q_n/n^s)$. Then $\zeta _{p*q}(s)$ is well-defined for $s\in {\mathbb {C}}_{>1}$ with $\zeta _{p*q}(s)=\zeta _p(s)\zeta _q(s)$. Since $p=I+p*q$, we have $\zeta _p(s)=\zeta _I(s)+\zeta _{p*q}(s)=1+\zeta _p(s)\zeta _q(s)~(s\in {\mathbb {C}}_{>1})$, so $\zeta _p(s)=(1-\zeta _q(s))^{-1}~(s\in {\mathbb {C}}_{>1})$. Let $f(x):=S_p(e^x)/e^x\in [0,2]~(x\ge 0)$. Then $\int _0^\infty f(x)e^{-(s-1)x}dx=\int _0^\infty S_p(e^x)e^{-sx}dx =\int _1^\infty S_p(x)/x^{s+1}dx=\zeta _p(s)/s=s^{-1}(1-\zeta _q(s))^{-1}~(s\in {\mathbb {C}}_{>1})$. Next, we define $F(s):=\int _0^\infty f(x)e^{-sx}dx~(s\in {\mathbb {C}}_{>0})$. Note that we have $F(s)=(1+s)^{-1}(1-\zeta _q(1+s))^{-1}~(s\in {\mathbb {C}}_{>0})$. Let $\sigma _n:=1/(n+2)~(n\ge 0)$. We would like to apply Theorem 2. Since $0\le f(x)\le 2$, we have $\int _0^\infty |f(x)|/(x+1)^2dx<+\infty$. Since $f(x)e^x=S_p(e^x)$ is non-decreasing, we have (1). Next, we show

• (a) $\forall u,v\in {\mathbb {R}}_{>0}~[0<u\le v~\Rightarrow ~ \lim _N\sup _{n,m\ge N}\int _u^v|\mathfrak {R}F(\sigma _n+it)-\mathfrak {R}F(\sigma _m+it)|dt=0]$,

• (b) $\exists T>0,~\exists K>0,~\exists n_0\ge 0,~\forall n\ge n_0,~a.e.t\in [0,T]~\left[ ~ \mathfrak {R}F(\sigma _n+it)\ge -K ~\right]$.

We first show (a). Since $F(s)=(1+s)^{-1}(1-\zeta _q(1+s))^{-1}~(s\in C_{>0})$ and $\zeta _q(s)$ is continuous on $s\in {\mathbb {C}}_{\ge 1}$ and $\zeta _q(s)\ne 1~(s\in {\mathbb {C}}_{\ge 1}-\left\{ 1 \right\} )$, we obtain (a) by an argument similar to the corresponding proof of Theorem 4. Next, we show (b). Since $\sum _{n=1}^\infty (q_n/n)=1$ and $q_1=0$, we have $q_M\ne 0$ for some $M\ge 2$. Let $C:=(1/4)\sum _{n=1}^M(q_n/n)\log ^2n$. Note that we have $0<C<(1/2)\sum _{n=1}^M(q_n/n)\log ^2n$. Then

$$\begin{aligned}&\liminf _{{\mathbb {R}}_{>0}^2\ni (\sigma ,t)\rightarrow (0,0)} \sum _{n=1}^\infty \frac{q_n}{n^{1+\sigma }}(\log ^2n)\mathrm{FK}(t\log n)\\&\quad \ge \liminf _{{\mathbb {R}}_{>0}^2\ni (\sigma ,t)\rightarrow (0,0)} \sum _{n=1}^M \frac{q_n}{n^{1+\sigma }}(\log ^2n)\mathrm{FK}(t\log n) =\frac{1}{2}\sum _{n=1}^M \frac{q_n}{n}\log ^2n>C. \end{aligned}$$

Then there exists $\delta >0$ such that $\sum _{n=1}^\infty (q_n/n^{1+\sigma })(\log ^2n)\mathrm{FK}(t\log n)>C$ for any $(\sigma ,t)\in (0,\delta ]^2$. Now let $(\sigma ,t)\in (0,\delta ]^2$ be arbitrary. Let $a=1-\sum _{n=1}^\infty (q_n/n^{1+\sigma })\cos (t\log n)$ and $b=\sum _{n=1}^\infty (q_n/n^{1+\sigma })\sin (t\log n)$. In view of $q_1=0$, we have

$$\begin{aligned} a&=1-\sum _{n=1}^\infty \frac{q_n}{n^{1+\sigma }}\cos (t\log n) =\sum _{n=2}^\infty \frac{q_n}{n}-\sum _{n=2}^\infty \frac{q_n}{n^{1+\sigma }}\cos (t\log n)\\&\ge \sum _{n=2}^\infty \frac{q_n}{n^{1+\sigma }}-\sum _{n=2}^\infty \frac{q_n}{n^{1+\sigma }}\cos (t\log n) =\sum _{n=2}^\infty \frac{q_n}{n^{1+\sigma }}(1-\cos (t\log n))\\&=t^2\sum _{n=2}^\infty \frac{q_n}{n^{1+\sigma }}(\log ^2n)\mathrm{FK}(t\log n)\ge Ct^2, \end{aligned}$$

i.e., we have $a\ge Ct^2$. Moreover, since $1-\zeta _q(1+\sigma +it)=a+ib$, we have

$$\begin{aligned} F(\sigma +it)&=\frac{1}{(1+\sigma +it)(1-\zeta _q(1+\sigma +it))}\\&=\frac{1}{(1+\sigma +it)(a+ib)}=\frac{(1+\sigma -it)(a-ib)}{((1+\sigma )^2+t^2)(a^2+b^2)}, \end{aligned}$$

so

$$\begin{aligned} \mathfrak {R}F(\sigma +it)=\frac{(1+\sigma )a-tb}{((1+\sigma )^2+t^2)(a^2+b^2)}=\frac{\eta a-tb}{(\eta ^2+t^2)(a^2+b^2)},~~\eta :=1+\sigma \ge 1. \end{aligned}$$

Since $a\ge Ct^2$, we have

$$\begin{aligned} \frac{\eta a-tb}{a^2+b^2}\ge \frac{\eta \cdot Ct^2-tb}{a^2+b^2}=\frac{a^2+(b-2C\eta t)^2}{4C\eta (a^2+b^2)}-\frac{1}{4C\eta }\ge \frac{-1}{4C\eta }, \end{aligned}$$

so

$$\begin{aligned} \mathfrak {R}F(\sigma +it)=\frac{\eta a-tb}{(\eta ^2+t^2)(a^2+b^2)}\ge \frac{1}{\eta ^2+t^2}\cdot \frac{-1}{4C\eta }\ge \frac{-1}{4C}. \end{aligned}$$

Thus we obtain $\mathfrak {R}F(\sigma +it)\ge -1/(4C)$ for any $(\sigma ,t)\in (0,\delta ]^2$. In particular, we obtain (b) with $T=\delta$, $K=1/(4C)$, and sufficiently large $n_0$. Then we have (i) and (ii) of Theorem 2 by renumbering the indices of $\sigma _{n+n_0}$. Thus we can apply Theorem 2, so $A:=\lim _{x\rightarrow +\infty }f(x)\in {\mathbb {R}}$ exists. Since $F(\sigma )=\int _0^\infty f(x)e^{-\sigma x}dx~(\sigma >0)$, we can easily show $\lim _{\sigma \downarrow 0}\sigma F(\sigma )=A$. Next, we show $\lim _{\sigma \downarrow 0}\sigma F(\sigma )=1/\sum _{n=1}^\infty (q_n/n)\log n$. Let $a(x):=(e^x-1)/x~(x\ne 0),~1~(x=0)$. Note that a(x) is non-decreasing on ${\mathbb {R}}$ and $a(x)\ge 0~(x\in {\mathbb {R}})$. Let $\sigma >0$. We can easily calculate

$$\begin{aligned} \frac{1-\zeta _q(1+\sigma )}{\sigma }=\cdots =\sum _{n=2}^\infty \frac{q_n\log n}{n}a(-\sigma \log n). \end{aligned}$$

Since $a(x)\ge 0$ is non-decreasing on ${\mathbb {R}}$ and $\lim _{x\rightarrow 0}a(x)=1$, for each $n\ge 2$ we have

$$\begin{aligned} 0\le \frac{q_n\log n}{n}a(-\sigma \log n)\uparrow \frac{q_n\log n}{n}~~(\sigma \downarrow 0). \end{aligned}$$

By the monotone convergence theorem (for series), we have $\lim _{\sigma \downarrow 0}(1-\zeta _q(1+\sigma ))/\sigma =\sum _{n=2}^\infty (q_n/n)\log n=\sum _{n=1}^\infty (q_n/n)\log n~(\in (0,+\infty ])$. In view of $F(\sigma )=(1+\sigma )^{-1}(1-\zeta _q(1+\sigma ))^{-1}~(\sigma >0)$, we have $\lim _{\sigma \downarrow 0}\sigma F(\sigma )=1/\sum _{n=1}^\infty (q_n/n)\log n$. Thus we obtain $A=1/\sum _{n=1}^\infty (q_n/n)\log n$, so $\lim _{x\rightarrow +\infty }f(x)=1/\sum _{n=1}^\infty (q_n/n)\log n$. Since $f(x)=S_p(e^x)/e^x=(1/e^x)\sum _{k\le e^x}p_k$, we obtain $\lim _n(1/n)\sum _{k=1}^np_k=1/\sum _{n=1}^\infty (q_n/n)\log n$. $\square$