# Some Properties of Weighted Pseudo almost Periodic Functions

Authors: Zhe-Ming Zheng; Hui-Sheng Ding; Gaston M. N’Guérékata
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Several interesting and new properties of weighted pseudo almost periodic functions are established. Firstly, we obtain an equivalent definition for weighted pseudo almost periodic functions, which shows a close relationship between asymptotically almost periodic functions and weighted pseudo almost periodic functions; secondly, we prove that the space of asymptotically almost periodic functions is always a proper subspace of the space of weighted pseudo almost periodic functions; thirdly, we show that under some cases, the space of weighted pseudo almost periodic functions equals the classical space of pseudo almost periodic functions.


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## Body

## 1. Introduction

The notion of weighted pseudo almost periodic function is introduced by Diagana [1], which is an interesting generalization of the classical Bohr almost periodic function as well as a generalization of the classical pseudo almost periodic function introduced by Zhang [2].Since the work of Diagana, it has been of great interest for many mathematicians to investigate weighted pseudo almost periodic functions and their applications to evolution equations. There is a large amount of literature on this topic. However, due to the influence of weighted term, the behaviors of weighted pseudo almost periodic functions are more tricky and changeable than those of the classical pseudo almost periodic functions.It is needed to note that, recently, several authors have made interesting and important contributions on weighted pseudo almost periodic functions. For example, Blot et al. [3, 4] established many results including completeness, translation invariance, and composition theorem of weighted pseudo almost periodic functions in the framework of measure theory; Diagana [5] investigated the weighted mean for almost periodic functions; Liang et al. [6] showed that the decomposition of weighted pseudo almost periodic functions is not unique, in general, and that translation invariance implies unique decomposition; Ji and Zhang [7] obtained several results concerning translation invariance and ergodicity of an almost periodic function under a weight; Zhang et al. [8] proved that the space of weighted pseudo almost periodic functions is complete under a new norm, which is different from the supremum norm; Ding et al. [9] presented several basic properties about vector-valued weighted pseudo almost automorphic functions, including equivalence, completeness, translation invariance, composition theorem, and convolution theorem of these functions. In addition, we would like to refer the reader to [10–13] for some other recent development on weighted pseudo almost periodic functions and related topics.Throughout the rest of this paper, we denote byℕ the set of positive integers, by ℝ the set of real numbers, by X a Banach space, by BC(ℝ,X) the set of all bounded continuous functions f:ℝ→X, and by 𝕌 the set of functions (weights) ρ:ℝ→[0,+∞), which are locally integrable over ℝ. In addition, for ρ∈𝕌 and r>0, we denote
(1)μ(r,ρ)∶=∫-rr‍ρ(t)dt,𝕌∞∶={ρ∈𝕌:limr→+∞μ(r,ρ)=+∞},𝕌B∶={ρ∈𝕌∞:ρisboundedwithinft∈ℝρ(t)>0}.
Obviously, 𝕌B⊂𝕌∞⊂𝕌, with strict inclusions.Next, let us recall some notions and basic results about almost periodic functions and pseudo almost periodic functions (for more details, see [14–16]).Definition 1.
A setP⊂ℝ is called relatively dense in ℝ if there exists a number l>0 such that, for all a∈ℝ, [a,a+l]⋂‍P≠⌀.Definition 2.
A continuous functionf:ℝ→X is called almost periodic if for every ɛ>0 there exists a relatively dense set P(ɛ) such that
(2)supt∈ℝ∥f(t+τ)-f(t)∥<ɛ
for all τ∈P(ɛ). We denote the set of all such functions by AP(X).Definition 3.
A continuous functionf:ℝ→X is called asymptotically almost periodic if it can be expressed as f=g+h, where g∈AP(X) and h∈C0(X), where
(3)C0(X)={h∈BC(ℝ,X):lim|t|→∞h(t)=0}.
The set of all such functions will be denoted by AAP(X).Lemma 4.
The following holds true:(a)
iff∈AAP(X), then f is uniformly continuous on ℝ;(b)
a necessary and sufficient condition for a continuous functionf:ℝ→X belonging to AAP(X) is that for every ɛ>0 there exist a constant T(ɛ)>0 and a relatively dense set P(ɛ) such that
(4)∥f(t+τ)-f(t)∥<ɛfor allt∈ℝ with |t|≥T(ɛ) and τ∈P(ɛ) with |t+τ|≥T(ɛ).Now, let us recall some notions and basic results about weighted pseudo almost periodic functions. Denote(5)PAP0(X,ρ)∶={f∈BC(ℝ,X):limr→+∞1μ(r,ρ)∫-rr‍∥f(t)∥ρ(t)dt=0}.Definition 5 (see [1]).
Letρ∈𝕌∞. A function f∈BC(ℝ,X) is called weighted pseudo almost periodic or ρ-pseudo almost periodic if it can be expressed as f=g+h, where g∈AP(X) and h∈PAP0(X,ρ). The set of such functions will be denoted by PAP(X,ρ).Remark 6.
Ifρ≡1, then a ρ-pseudo almost periodic function becomes a classical pseudo almost periodic function. So PAP(X,1) is just the space of all pseudo almost periodic functions. In addition, it is easy to show that PAP(X,ρ)=PAP(X,1) provided that ρ∈𝕌B.Definition 7 (see [7]).
Letρ∈𝕌∞. A set C⊂ℝ is said to be a ρ-ergodic zero set if
(6)limr→+∞∫[-r,r]∩C‍ρ(t)dtμ(r,ρ)=0.The following lemma is due to [17, Lemma  3.2] (see also [7]).Lemma 8.
Letf∈BC(ℝ,X) and ρ∈𝕌∞. Then f∈PAP0(X,ρ) if and only if, for every ɛ>0, Mɛ(f) is a ρ-ergodic zero set, where Mɛ(f)∶={t∈ℝ:∥f(t)∥≥ɛ}.


## 2. The Relationship between AAP(X) and PAP(X,ρ)

In this section, we discuss the relationship betweenAAP(X) and PAP(X,ρ). Firstly, we will establish an equivalent definition for weighted pseudo almost periodic functions.Theorem 9.
Letf∈BC(ℝ,X) and ρ∈𝕌∞. Then a necessary and sufficient condition for f∈PAP0(X,ρ) is that there exists a ρ-ergodic zero set C⊂ℝ such that
(7)lim|t|→∞t∈ℝ∖Cf(t)=0.Proof.
Consider the following.
Sufficiency. It follows from lim|t|→∞t∈ℝ∖Cf(t)=0 that, for every ɛ>0,
(8){t∈ℝ∖C:∥f(t)∥≥ɛ}
is a bounded set. Combining this with the fact that C is a ρ-ergodic zero set, we conclude that Mɛ(f)={t∈ℝ:∥f(t)∥≥ɛ} is a ρ-ergodic zero set. Then, by Lemma 8, f∈PAP0(X,ρ).
Necessity. Let f∈PAP0(X,ρ). We denote
(9)Cn={t∈ℝ:∥f(t)∥>1n},n∈ℕ.
Then, by Lemma 8, every Cn is a ρ-ergodic zero set. Next, we divide the remaining proof into three steps.
Step  1. There exists an increasing sequence {Tn}n=1∞⊂ℝ with limn→∞Tn=∞, satisfying that, for every n∈ℕ and r≥Tn, there hold
(10)∫[-r,r]∩Cn+1‍ρ(t)dtμ(r,ρ)≤1n,∑i=1n∫[-Ti,Ti]∩Ci‍ρ(t)dtμ(r,ρ)≤1n.
In fact, forn=1, since C1 and C2 are both ρ-ergodic zero sets, there exists T1>1 such that, for all r≥T1, there hold
(11)∫[-r,r]∩C2‍ρ(t)dtμ(r,ρ)≤1,∫[-T1,T1]∩C1‍ρ(t)dtμ(r,ρ)≤∫[-r,r]∩C1‍ρ(t)dtμ(r,ρ)≤1.
For n=2, by using the fact that C2 and C3 are both ρ-ergodic zero sets, there exists T2>max{T1,2} such that, for all r≥T2, there hold
(12)∫[-r,r]∩C3‍ρ(t)dtμ(r,ρ)≤12,∑i=12∫[-Ti,Ti]∩Ci‍ρ(t)dtμ(r,ρ)≤∫[-T1,T1]∩C1‍ρ(t)dt+∫[-r,r]∩C2‍ρ(t)dtμ(r,ρ)≤12.
Forn=k, by using the fact that Ck and Ck+1 are both ρ-ergodic zero, there exists Tk>max{Tk-1,k} such that, for all r≥Tk, there hold
(13)∫[-r,r]∩Ck+1‍ρ(t)dtμ(r,ρ)≤1k,∑i=1k∫[-Ti,Ti]∩Ci‍ρ(t)dtμ(r,ρ)≤∑i=1k-1∫[-Ti,Ti]∩Ci‍ρ(t)dt+∫[-r,r]∩Ck‍ρ(t)dtμ(r,ρ)≤1k.
Continuing by this way, we can get an increasing sequence {Tn}n=1∞, which satisfies limn→∞Tn=∞, (10).
Step  2.  C∶=⋃n=1∞C¯n is a ρ-ergodic zero set, where
(14)C¯1=[-T1,T1]∩C1,C¯n=([-Tn,Tn]∖[-Tn-1,Tn-1])∩Cn,hhhhhhhhhhhhiiiiiiiin=2,3,….
In fact, for everyr>0, there exists Tn such that r∈[Tn,Tn+1). Thus, we have
(15)[-r,r]⊂[-T1,T1]∪([-T2,T2]∖[-T1,T1])∪⋯([-Tn,Tn]∖[-Tn-1,Tn-1])∪([-r,r]∖[-Tn,Tn]),([-r,r]∩C)⊂C¯1∪⋯C¯n∪([-r,r]∩C¯n+1).
Combining this with (10), we conclude that
(16)∫[-r,r]∩C‍ρ(t)dtμ(r,ρ)≤∑i=1n∫C¯i‍ρ(t)dtμ(r,ρ)+∫[-r,r]∩C¯n+1‍ρ(t)dtμ(r,ρ)≤∑i=1n∫[-Ti,Ti]∩Ci‍ρ(t)dtμ(r,ρ)+∫[-r,r]∩Cn+1‍ρ(t)dtμ(r,ρ)≤2n,
which means that C is a ρ-ergodic zero set.
Step  3. lim|t|→∞t∈ℝ∖Cf(t)=0.
In fact, for everyn∈ℕ and t∈ℝ∖C with |t|>Tn, there exists k≥n such that
(17)t∈[-Tk+1,Tk+1]∖[-Tk,Tk],
which yields that t∉Ck+1, and thus
(18)∥f(t)∥<1k+1<1n.
This completes the proof.Remark 10.
In Theorem9, we can choose C to be an open set. In fact, letting
(19)C¯=C⋃‍(⋃n=1∞‍(Tn,Tn+δn))⋃‍(⋃n=1∞‍(-Tn-δn,-Tn)),
where every δn is a sufficiently small positive constant satisfying
(20)∫TnTn+δn‍ρ(t)dt+∫-Tn-δn-Tn‍ρ(t)dt<12n,n∈ℕ,
it is easy to prove that C¯ is an open set and a ρ-ergodic zero set. Moreover, obviously, we have lim|t|→∞t∈ℝ∖C¯f(t)=0.Next, we present an equivalent definition of weighted pseudo almost periodic functions, which establishes a close relationship between asymptotically almost periodic functions and weighted pseudo almost periodic functions.Theorem 11.
Letf∈BC(ℝ,X) and ρ∈𝕌∞. Then a necessary and sufficient condition for f∈PAP(X,ρ) is that there exist f¯∈AAP(X) and a ρ-ergodic zero set C⊂ℝ such that
(21)f¯(t)=f(t),t∈ℝ∖C.Proof.
The sufficiency part is easy to prove. We only give the proof for the necessity part. Letf∈PAP(X,ρ). Then, there exist g∈AP(X) and h∈PAP0(X,ρ) such that f=g+h. It follows from Theorem 9 that there exists a ρ-ergodic zero set C⊂ℝ such that
(22)lim|t|→∞t∈ℝ∖Ch(t)=0.
In addition, by Remark 10, without loss of generality, we can assume that C is an open set. Then, we can conclude that there exists a function h¯∈C0(X) such that h¯(t)=h(t) for all t∈ℝ∖C. Letting f¯=g+h¯, we have f¯∈AAP(X), and f¯(t)=f(t) for all t∈ℝ∖C. This completes the proof.Combining Lemma4 and Theorem 11, we can get the following.Corollary 12.
Letρ∈𝕌∞ and f∈PAP(X,ρ). Then, there exists a ρ-ergodic zero set C⊂ℝ satisfying that, for every ɛ>0, there are two constants δ(ɛ),T(ɛ)>0 and a relatively dense set P(ɛ) such that(23)∥f(t+τ)-f(t)∥<ɛ,τ∈P(ɛ),t,t+τ∉C∪[-T(ɛ),T(ɛ)],∥f(t1)-f(t2)∥<ɛ,t1,t2∉C,|t1-t2|<δ(ɛ).Remark 13.
For the case ofρ≡1, Zhang [18] established a similar result to Corollary 12. But here we use a different approach, and even for the case of ρ≡1, Corollary 12 improves the “if part” of [18, Theorem  11]. In fact, in [18], for every ɛ>0, there exists a 1-ergodic zero set C(ɛ). Here, we find a common ergodic zero set C.In the above, we show that there is a close relationship between asymptotically almost periodic functions and weighted pseudo almost periodic functions. Next, we will show thatAAP(X) is always a proper subspace of PAP(X,ρ).Theorem 14.
Letρ∈𝕌∞. Then
(24)AAP(X)⫋PAP(X,ρ).Proof.
It is easy to show thatAAP(X)⊂PAP(X,ρ) since C0(X)⊂PAP0(X,ρ). So we only need to show that AAP(X)≠PAP(X,ρ). Without loss for generality, we only give the proof for the case of X=ℝ. We divide the remaining proof into three steps.
Step  1. Let C=⋃n=1∞Cn, where
(25)Cn:={t∈(n-1,n):F(t,n)<1nF(n-1,n)},n∈ℕ,
where
(26)F(x,y)∶=∫A(x,y)‍ρ(s)ds,A(x,y)∶={s∈ℝ:x≤|s|≤y}hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhy≥x≥0.
It is easy to see that every Cn is open. Moreover, there are infinitely many nonempty Cn, and thus we can assume that every Cn is nonempty without loss for generality. In addition, we claim that C is a ρ-ergodic zero set. In fact, we have the following two cases.
Case  I. If r∈C, then r∈C[r]+1, which yields that
(27)∫[-r,r]∩C‍ρ(t)dtμ(r,ρ)≤∫[-r,r]∩C‍ρ(t)dt+∫A(r,[r]+1)‍ρ(t)dt∫[-r,r]‍ρ(t)dt+∫A(r,[r]+1)‍ρ(t)dt=∫[-r,r]∩C‍ρ(t)dt+F(r,[r]+1)μ([r]+1,ρ)≤∫[-r,r]∩C‍ρ(t)dt+(1/([r]+1))F([r],[r]+1)μ([r]+1,ρ)≤∑k=1[r]+1∫A(k-1,k)∩C‍ρ(t)dt+(1/([r]+1))F([r],[r]+1)μ([r]+1,ρ)≤∑k=1[r]+1∫Ck‍ρ(t)dt+(1/([r]+1))F([r],[r]+1)μ([r]+1,ρ)≤∑k=1[r]+1(1/k)F(k-1,k)+(1/([r]+1))F([r],[r]+1)μ([r]+1,ρ)=∑k=1[r](1/k)F(k-1,k)+(2/([r]+1))F([r],[r]+1)μ([r]+1,ρ)=(∑k=1m1kF(k-1,k)+∑k=m+1[r]1kF(k-1,k)+2[r]+1F([r],[r]+1)∑k=1m(1/k)F(k-1,k))hhh×(μ([r]+1,ρ))-1≤μ(m,ρ)+(2/(m+1))μ([r]+1,ρ)μ([r]+1,ρ)≤μ(m,ρ)μ([r]+1,ρ)+2m+1,
where m<[r]-1 is a positive integer. Then, we get
(28)limr→+∞∫[-r,r]∩C‍ρ(t)dtμ(r,ρ)=0.
Case  II. If r∉C, then ([r],r]∩C=∅, which yields that
(29)∫[-r,r]∩C‍ρ(t)dtμ(r,ρ)≤∑k=1[r]∫A(k-1,k)∩C‍ρ(t)dt+∫A([r],r)∩C‍ρ(t)dtμ(r,ρ)=∑k=1[r]∫A(k-1,k)∩C‍ρ(t)dtμ(r,ρ).
Then, similar to Case  I, one can also obtain that
(30)limr→+∞∫[-r,r]∩C‍ρ(t)dtμ(r,ρ)=0.
Thus, C is a ρ-ergodic zero set.
Step  2. For every n∈ℕ, noting that Cn is open, there exist tn∈Cn and δn>0 such that (tn-δn,tn+δn)⊂Cn. Now, we construct a bounded and continuous function on ℝ by
(31)φ(t)={t-tn(n+1)/2+δn(n+1)/2δn(n+1)/2,t∈[tn(n+1)/2-δn(n+1)/2,tn(n+1)/2],n∈ℕ,tn(n+1)/2+δn(n+1)/2-tδn(n+1)/2,t∈[tn(n+1)/2,tn(n+1)/2+δn(n+1)/2],n∈ℕ,0,otherwise.
Step  3. φ∈PAP(X,ρ)∖AAP(X).
Since{t∈ℝ:φ(t)≠0}⊂C and C is a ρ-ergodic zero set, we have φ∈PAP0(X,ρ)⊂PAP(X,ρ). It remains to show that φ∉AAP(X). We prove it by contradiction, assuming that there exist φ1∈AP(X) and φ2∈C0(X) such that φ=φ1+φ2. For sufficiently large n, since φ1∈AP(X), we can choose
(32)τn∈[n(n+1)2-tn(n+1)/2,(n+1)(n+2)2-1-tn(n+1)/2]
such that
(33)|φ1(tn(n+1)/2+τn)-φ1(tn(n+1)/2)|<12.
Moreover, since φ2∈C0(X), for sufficiently large n, we also have
(34)|φ2(tn(n+1)/2+τn)-φ2(tn(n+1)/2)|<12.
So, we get
(35)|φ(tn(n+1)/2+τn)-φ(tn(n+1)/2)|<1,
which contradicts the fact that
(36)φ(tn(n+1)/2)=1,φ(tn(n+1)/2+τn)=0.
This completes the proof.


## 3. Equivalence

Just as noted in Remark6, we know that PAP(X,ρ)=PAP(X,1) provided that ρ∈𝕌B. Then, there is a natural question:DoesPAP(X,ρ)=PAP(X,1) imply that ρ∈𝕌B?In fact, the above question has a negative answer. For example, recently, it is proved in [7] and [9] (by a different method) that
(37)PAP(X,|·|n)=PAP(X,1),n∈ℕ.In this section, we will make further study on this question. We will prove that for some otherρ∉𝕌B, there still holds PAP(X,ρ)=PAP(X,1). Firstly, we recall a theorem, which is due to [7, Theorem  4.3].Theorem 15.
Letρ1,ρ2∈𝕌∞ and
(38){C⊂ℝ:Cisaρ1-ergodiczeroset}={C⊂ℝ:Cisaρ2-ergodiczeroset}.
Then PAP0(X,ρ1)=PAP0(X,ρ2).Theorem 16.
Letρ∈𝕌∞ be a periodic function with ρ(t)>0 almost everywhere on ℝ. Then PAP0(X,ρ)=PAP0(X,1), and thus PAP(X,ρ)=PAP(X,1).Proof.
It suffices to prove thatPAP0(X,ρ)=PAP0(X,1). We divide the remaining proof into two steps.
Step  1. Every ρ-ergodic zero set is a 1-ergodic zero set.
LetC be a ρ-ergodic zero set. Then, we have
(39)limr→+∞∫[-r,r]∩C‍ρ(t)dtμ(r,ρ)=0.
Assuming that
(40)limr→+∞mes([-r,r]∩C)2r≠0,
then there exist ɛ0>0 and a sequence of positive numbers {Tn}n=1∞, which satisfies limn→∞Tn=∞, and
(41)mes([-Tn,Tn]∩C)>2Tnω·ɛ0,
where ω is a positive periodic of ρ. On the other hand, since ρ(t)>0 almost everywhere on ℝ, by Lusin's Theorem, there exists a closed set F⊂[0,ω] with mes([0,ω]-F)<ɛ0/2, such that ρ is continuous on F and ρ(t)>0 for all t∈F. Let
(42)m0∶=inft∈Fρ(t).
Then m0>0. Then, we have
(43)mes{t∈[0,ω]:ρ(t)<m0}<ɛ02,
which yields that
(44)mes{t∈[-Tn,Tn]:ρ(t)<m0}<([2Tnω]+1)·ɛ02.
Combining this with (41), we get
(45)∫[-Tn,Tn]∩C‍ρ(t)dtμ(Tn,ρ)≥m0·{(2Tn/ω)·ɛ0-([2Tn/ω]+1)·(ɛ0/2)}∫[-Tn,Tn]‍ρ(t)dt≥m0·{(2Tn/ω)·ɛ0-([2Tn/ω]+1)·(ɛ0/2)}([2Tn/ω]+1)·∫0ω‍ρ(t)dt⟶m0·(ɛ0/2)∫0ω‍ρ(t)dt,n⟶∞,
which contradicts (39). Therefore, C is a 1-ergodic zero set.
Step  2. Every 1-ergodic zero set is a ρ-ergodic zero set.
LetC be a 1-ergodic zero set. For every ɛ>0, there exists δ>0 such that, for all set E⊂[0,ω] with mesE<δ, there holds
(46)∫E‍ρ(t)dt<ɛ.
In addition, again by Lusin's Theorem, for the above δ>0, there exists a closed set Fδ⊂[0,ω] with mes([0,ω]-Fδ)<δ, such that ρ is continuous on Fδ. Let
(47)Mδ=supt∈Fδρ(t).
Then Mδ∈(0,+∞). Recalling that C is a 1-ergodic zero set, for the above ɛ and Mδ, there exists rɛ>0 such that
(48)limr→+∞mes([-r,r]∩C)+2ω2r<ɛMδ,r≥rɛ.
Combining (46) and (48), by using the periodicity of ρ, we conclude that, for all r≥rɛ, there holds
(49)∫[-r,r]∩C‍ρ(t)dt≤∑k=0[r/ω]‍∫[kω,kω+ω]∩C‍ρ(t)dt+∑k=0[r/ω]‍∫[-kω-ω,-kω]∩C‍ρ(t)dt≤([rω]+1)·2ɛ+Mδ·(mes([-r,r]∩C)+2ω)≤([rω]+1)·2ɛ+2rɛ,
where
(50)∑k=0[r/ω]‍∫[kω,kω+ω]∩C‍ρ(t)dt≤∑k=0[r/ω]‍∫[[kω,kω+ω]-(Fδ+kω)]∩C‍ρ(t)dt+∑k=0[r/ω]‍∫[[kω,kω+ω]∩(Fδ+kω)]∩C‍ρ(t)dt≤∑k=0[r/ω]‍∫[kω,kω+ω]-(Fδ+kω)‍ρ(t)dt+∑k=0[r/ω]‍Mδ·mes([kω,kω+ω]∩C)≤∑k=0[r/ω]‍∫[0,ω]-Fδ‍ρ(t)dt+Mδ·(mes([0,r]∩C)+ω)≤([rω]+1)·ɛ+Mδ·(mes([0,r]∩C)+ω),
and, similarly,
(51)∑k=0[r/ω]‍∫[-kω-ω,-kω]∩C‍ρ(t)dt≤([rω]+1)·ɛ+Mδ·(mes([-r,0]∩C)+ω).
Noting that
(52)∫-rr‍ρ(t)dt≥2[rω]·∫0ω‍ρ(t),
we get
(53)∫[-r,r]∩C‍ρ(t)dt∫-rr‍ρ(t)dt≤([r/ω]+1)·2ɛ+2rɛ2[r/ω]·∫0ω‍ρ(t)dt⟶ω+1∫0ω‍ρ(t)dt·ɛ,r⟶+∞.
Then, by the arbitrariness of ɛ, we conclude that
(54)limr→+∞∫[-r,r]∩C‍ρ(t)dt∫-rr‍ρ(t)dt=0;
that is, C is a ρ-ergodic zero set.Example 17.
Letρ(t)=cos2n(t), where n∈ℕ. Then by Theorem 16, PAP(X,ρ)=PAP(X,1).Except for the case in Theorem16, there is some other ρ, which satisfies that PAP0(X,ρ)=PAP0(X,1). For example, we have the following.Theorem 18.
Let(55)ρ(t)={ln|t|,|t|>1,0,|t|≤1.
Then PAP0(X,ρ)=PAP0(X,1).Proof.
We first show that every1-ergodic zero set is a ρ-ergodic zero set. Let C be a 1-ergodic zero set. For every ɛ>0, there exists R>0 such that, for all r>R,
(56)mes([-r,r]∩C)<2r·ɛ.
Combining this with the fact that ρ is even and increasing on [0,+∞), we get
(57)μ([-r,r]∩C,ρ)μ([-r,r],ρ)⩽2rlnr·ɛ2(rlnr-r+1),r>R,
which yields that
(58)limsupr→+∞μ([-r,r]∩C,ρ)μ([-r,r],ρ)≤ɛ.
Then, by the arbitrariness of ɛ, we know that
(59)limr→+∞μ([-r,r]∩C,ρ)μ([-r,r],ρ)=0;
that is, C is a ρ-ergodic zero set.
Next, let us show that everyρ-ergodic zero set is a 1-ergodic zero set. Let C be a ρ-ergodic zero set. Assuming that
(60)limr→+∞mes([-r,r]∩C)2r≠0
then there exist ɛ0>0 and a sequence of positive numbers {Tn}n=1∞, which satisfies limn→∞Tn=∞, and
(61)mes([-Tn,Tn]∩C)>2Tnɛ0.
By (61) and the fact that ρ is even and increasing on [0,+∞), we conclude that
(62)μ([-Tn,Tn]∩C,ρ)μ([-Tn,Tn],ρ)⩾μ([-Tnɛ0,Tnɛ0],ρ)μ([-Tn,Tn],ρ)=ɛ0(Tnln(ɛ0Tn)-Tn)+1TnlnTn-Tn+1=ɛ0(TnlnTn-Tn)+Tnɛ0lnɛ0+1TnlnTn-Tn+1.
Letting n→∞, we get
(63)liminfn→∞μ([-Tn,Tn]∩C,ρ)μ([-Tn,Tn],ρ)≥ɛ0,
which contradicts the fact that C is a ρ-ergodic zero set. Thus, C is a 1-ergodic zero set. This completes the proof.



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*Source: 327213-2013-12-09.xml*