# The Hyperorder of Solutions of Second-Order Linear Differential Equations

Authors: Guowei Zhang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We prove that the hyperorder of every nontrivial solution of homogenous linear differential equations of typef′′+A1(z)eazf′+A0(z)ebzf=0 and nonhomogeneous equation of type f′′+A1(z)eazf′+A0(z)ebzf=H(z) is one, where A0,A1,H(z) are entire functions of order less than one, improving the previous results of Chen, Wang, and Laine.


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## Body

## 1. Introduction

We assume that the reader is familiar with the usual notations and the basic results of the Nevanlinna theory (see [1–4]). We also use basic notions and the results of the Wiman-Valiron theory; see [5]. Let f(z) be a nonconstant meromorphic function in the complex plane. We remark that σ(f), respectively, σ2(f) will be used to denote the order, respectively, the hyperorder, of f. In particular, the hyperorder σ2(f) is defined as
(1)σ2(f)=limsupr→∞loglogT(r,f)logr;
see [1, 2, 4]. For a set E⊂R+, let m(E), respectively, λ(E), denote the linear measure, respectively, the logarithmic measure of E. Moreover, the upper logarithmic density and the lower logarithmic density of E are defined by
(2)logdens¯(E)=limsupr→∞λ(E⋂‍[1,r])logr,logdens_(E)=liminfr→∞λ(E⋂‍[1,r])logr.
Observe that E may have a different meaning at different occurrences in what follows.We now recall some previous results concerning linear differential equations of type(3)f′′+A1(z)eazf′+A0(z)ebzf=0,(4)f′′+A1(z)eazf′+A0(z)ebzf=H(z),
where A0,A1,H(z) are entire functions of order less than one, and a,b are complex constants.Chen proved the following theorem; see [6].Theorem A.
LetA0≢0, A1≢0 be entire functions of order less than one, and the complex constants a,b satisfy ab≠0 and a=cb(c>1). Then every nontrivial solution f of (3) is of infinite order.Wang and Laine investigated the nonhomogeneous equation (4) and got the following; see [7].Theorem B.
Suppose thatA0≢0, A1≢0, H are entire functions of order less than one, and the complex constants a,b satisfy ab≠0 and b≠a. Then every nontrivial solution f of (4) is of infinite order.Theorem C.
Suppose thatA0≢0, A1≢0, D0, D1, H are entire functions of order less than one, and the complex constants a,b satisfy ab≠0 and b/a<0. Then every nontrivial solution f of equation
(5)f′′+(A1(z)eaz+D1(z))f′+(A0(z)ebz+D0(z))f=H(z)
is of infinite order.In this paper, we investigate the hyperorder of the nontrivial solutions of (3), (4), and (5) and obtain the following theorems.Theorem 1.
Suppose thatA0≢0, A1≢0, H are entire functions of order less than one, and the complex constants a,b satisfy ab≠0 and b≠a. Then the hyperorder of every nontrivial solution f of (4) is one.Corollary 2.
LetA0≢0, A1≢0 be entire functions of order less than one, and the complex constants a,b satisfy ab≠0 and a=cb(c>1). Then the hyperorder of every nontrivial solution f of (3) is one.Theorem 3.
Suppose thatA0≢0, A1≢0, D0, D1, H are entire functions of order less than one, and the complex constants a,b satisfy ab≠0 and b/a<0. Then the hyperorder of every nontrivial solution f of (5) is one.


## 2. Lemmas

Lemma 4 (see [5]).
Letf be an entire function of infinite order and let νf(r) be the central index of f(z), then the hyperorder
(6)σ2(f)=limsupr→∞loglogνf(r)logr.Lemma 5 (see [8]).
Letf be an entire function of infinite order with σ2(f)=α(0≤α<∞), and there exists a set E2⊂[1,∞) which have a finite logarithmic measure. Then there exists a sequence {zn=rneiθn} such that |f(zn)|=M(rn,f), θn∈[0,2π), limn→∞θn=θ0∈[0,2π), and rn∉E2, rn→∞ and such that(1)
ifσ2(f)=α(0<α<∞), then, for any given ɛ1(0<ɛ1<α),
(7)exp{rnα-ɛ1}<ν(rn)<exp{rnα+ɛ1};(2)
ifσ2(f)=0, then, for any given ɛ2(0<ɛ2<1/2) and for any large M1(>0),
(8)rnM1<ν(rn)<exp{rnɛ2}.Lemma 6 (see [7]).
Suppose thatP(z)=(α+iβ)z, where α,β are real numbers, |α|+|β|≠0, and that A(z)(≢0) is a meromorphic function with σ(A)<1. Set g(z)=A(z)eP(z), z=reiθ, δ(P,θ)=αcosθ-βsinθ. Then, for any given ɛ>0, there exists a set E3⊂(1,∞) of finite linear measure such that, for any θ∈[0,2π)∖H, there exists R>0 such that, for |z|=r>R and r∉E3, we have(1)
ifδ(P,θ)>0, then
(9)exp{(1-ɛ)δ(P,θ)r}<|g(reiθ)|<exp{(1+ɛ)δ(P,θ)r};(2)
ifδ(P,θ)<0, then
(10)exp{(1+ɛ)δ(P,θ)r}<|g(reiθ)|<exp{(1-ɛ)δ(P,θ)r},where H={θ∈[0,2π)∣δ(P,θ)=0}.Lemma 7 (see [6]).
LetA,B be entire functions of finite order and if f is a solution of equation
(11)f′′+Af′+Bf=0,
then the hyperorder σ2(f)≤max{σ(A),σ(B)}.The proof of the lemma below follows the idea of Bergweiler et al.; see [9, Theorem 3.1].Lemma 8.
Letf(z) be an entire function, and M(r,f)=f(reiθr) for every r. Set θr→θ0, and there exists a constant l0>0 and a set E with positive lower logarithmic density such that
(12)M(r,f)1/5≤|f(reiθ)|
for all r∈E large enough and all θ such that |θ-θ0|<l0.Proof.
Sincef is entire function, we know that M(r,f) is nondecrease, M(r,f)→∞ as r→∞, and |f(reiθ)| is continuous on the circle |z|=r for z=reiθ. Set θr→θ0∈[0,2π) as r→∞. A(ua,ub) denotes an annulus for 0<a<b and sufficiently large u. Then, there exists a constant 4l0(<π) such that |f(reiθ)|>1 for z=reiθ∈D, where D:={(r,θ)∣r∈A(ua,ub),|θ-θ0|<4l0} for u sufficiently large. Then the function h(z):=log|f(z)| is a positive harmonic in D. So H(t)=h(et) is a positive harmonic in the domain S:={t∣alogu<ℜ(t)<blogu,θ0-4l0<ℑ(t)<θ0+4l0}. Thus, if t1 and t2 satisfy alogu+3l0<ℜ(t1)=ℜ(t2)<blogu-3l0 and |ℑ(t1)-ℑ(t2)|<2l0, where ℑ(tj)∈(θ0-l0,θ0+l0), j=1,2, then {t∣|t-t1|<3l0} and |t1-t2|<2l0<3l0. So
(13)15=3l0-2l03l0+2l0≤H(t2)H(t1)≤3l0+2l03l0-2l0=5
by Harnack's inequality; see [10, Theorem 1.3.1]. Therefore, if z1 and z2 are in the domain D1:={(r,θ)∣r∈A(ua,ub),|θ-θ0|<l0}, where u is sufficiently large, then
(14)15≤h(z2)h(z1)≤5.
Set m*(r,f)=min|z|=r,|θ-θ0|<l0|f(reiθ)|. Then, we have 1/5logM(r,f)≤logm*(r,f) for z∈D1. If let u→∞, and then the set E of r∈(ua,ub) is of positive lower logarithmic density. Thus, the conclusion of this lemma holds.Lemma 9.
Letf(z) be an entire function with infinite order and let hyperorder σ2(f)≤1, g(z) be an entire function with finite order σ(g)<∞. For r∈E, where E is the infinite logarithmic measure set which is given in Lemma 8. Then, for any given ɛ0,
(15)|g(z)f(z)|<ɛ0
for all z such that |z|=r∈E is sufficiently large and that |f(z)|=M(r,f).Proof.
Since, for the entire functiong(z),
(16)σ(g)=limsupr→∞loglogM(r,g)logr
for any given ɛ, we have
(17)|g(z)|≤exp{rσ(g)+ɛ}
for all r sufficiently large. Since the order of f is infinite, for r∈E, there exists a sufficiently large real number A such that
(18)σ(f)=limsupr→∞loglogM(r,f)logr>A.
Thus, for r∈E is sufficiently large,
(19)M(r,f)≥exp{rA-ɛ}.
By (17) and (19), we conclude that
(20)|g(z)f(z)|≤exp{rσ(g)+ɛ}M(r,f)≤exp{rσ(g)+ɛ}exp{rA-ɛ}⟶0
for all z satisfying |f(z)|=M(r,f) such that r∈E is sufficiently large. Thus, the conclusion holds.


## 3. Proofs of Theorems

Proof of Theorem1.
Suppose thatf is a solution of (4), and then f is an entire function.
Step  1. We prove that σ2(f)≤1. Since σ(A0;A1;H)<1, set σ(H)=λ<1. Then for any given ɛ satisfying ɛ<1-λ, when r is sufficiently large, we have
(21)|A1eaz|≤exp{r1+ɛ},(22)|A0ebz|≤exp{r1+ɛ},(23)|H(z)|≤exp{rλ+ɛ}.
From the Wiman-Valiron theory, there is a set E1 having finite logarithmic measure, such that
(24)f(j)(z)f(z)=(νf(r)z)j(1+O(1))(j=1,2)
whenever |f(z)|=M(r,f),r∉E1, where the νf(r) is the central index of f(z), and we know that νf(r)→∞ as r→∞. When r sufficiently large, we have |f(z)|=M(r,f)>1. From (4) we have
(25)|f′′f|≤|A1eaz||f′f|+|A0ebz|+|Hf|.
Substituting (21), (22), (23), and (24) into (25), we obtain
(26)(νf(r)|z|)2(1+O(1))≤exp{r1+ɛ}νf(r)|z|(1+O(1))+exp{r1+ɛ}+exp{rλ+ɛ},
where z satisfies |z|=r∉E1 and r sufficiently large. By (26) we get
(27)limsupr→∞loglogνf(r)logr≤1+ɛ.
Since ɛ is arbitrary, by (27) and Lemma 4, we have σ2(f)≤1.
Step  2. By Theorem B, we know that the order of f is infinite, and, by the first step, we clear that the hyperorder of f is less than one. Thus, by Lemma 9 and (23), we have
(28)|Hf|<ɛ0
for all z satisfying |f(z)|=M(r,f) such that r∈E is sufficiently large, where E is of infinite logarithmic measure. Set σ2(f)=α0, and we assert that α0=1. Now we assume that α0<1, and prove that σ2(f)=α0<1 results in contradictions. E2,E3 are the sets in Lemmas 5 and 6, respectively.
Sinceλ(E1∪E2∪E3)<∞, we have that λ(E∖(E1∪E2∪E3)) is infinite. Thus, by Lemma 5, we see that there exists a sequence of points {zn=rneiθn} such that |f(zn)|=M(rn,f), θn∈[0,2π), limn→∞θn=θ0∈[0,2π), rn∈E∖(E1∪E2∪E3), rn→∞, and if σ2(f)=α0(0<α0<∞), then, for any given ɛ1(0<ɛ1<min{α0,1-α0}),
(29)exp{rnα0-ɛ1}<νf(rn)<exp{rnα0+ɛ1};
if σ2(f)=0, then, for any given ɛ2(0<ɛ2<1/2) and for any large M1(>0),
(30)rnM1<νf(rn)<exp{rnɛ2}.
Firstly, we prove the case when σ2(f)=α0(>0). It can separate into three cases to discuss.
Case  1. First assume that δ(az,θ0)>0. From the continuity of δ(az,θ0), we have
(31)12δ(az,θ0)<δ(az,θn)<32δ(az,θ0)
for sufficiently large n. From (9), we deduce that
(32)exp{1-ɛ2δ(az,θ0)rn}≤|A1(zn)eazn|≤exp{3(1+ɛ)2δ(az,θ0)rn}
for all n sufficiently large.
From (4), we have
(33)|f′(zn)f(zn)+A0(zn)A1(zn)e(b-a)zn|≤1|A1(zn)eazn|(|f′′(zn)f(zn)|+|H(zn)f(zn)|).
Subcase  1.1.We first assume that θ0 satisfies η:=δ((b-a)z,θ0)>0. From the continuity of δ((b-a)z,θ), we also have
(34)exp{1-ɛ2ηrn}≤|A0(zn)A1(zn)e(b-a)zn|≤exp{3(1+ɛ)2ηrn}
for all n sufficiently large. From (33), we get
(35)|A0(zn)A1(zn)e(b-a)zn|≤|f′(zn)f(zn)|+1|A1(zn)eazn|(|f′′(zn)f(zn)|+|H(zn)f(zn)|).
Substituting  (24), (28), (29), (32), and (34) into (35), we obtain
(36)exp{1-ɛ2ηrn}≤exp{rnα0+ɛ1}rn-1(1+O(1))+exp{-1-ɛ2δ(az,θ0)rn}×(exp{2rnα0+ɛ1}rn-2(1+O(1))+ɛ0).
Since α0+ɛ1<1, we see that (36) are contradictory as n→∞.
Subcase  1.2. Next assume that η:=δ((b-a)z,θ0)<0. Then, from (10), for n large enough, we deduce that
(37)exp{3(1+ɛ)2ηrn}≤|A0(zn)A1(zn)e(b-a)zn|≤exp{1-ɛ2ηrn}.
From (33), we get
(38)|f′(zn)f(zn)|≤|A0(zn)A1(zn)e(b-a)zn|+1|A1(zn)eazn|(|f′′(zn)f(zn)|+|H(zn)f(zn)|).
Substituting  (24), (28), (29), (32), and (37) into (38), we obtain
(39)νf(rn)rn(1+O(1))≤exp{1-ɛ2ηrn}+exp{-1-ɛ2δ(az,θ0)rn}×(exp{2rnα0+ɛ1}rn-2(1+O(1))+ɛ0)
as n→∞. Since α0+ɛ1<1, this implies that νf(r)→0, n→∞, which is impossible.
Subcase  1.3. Assume finally that η:=δ((b-a)z,θ0)=0. Here, (12) may be used to construct another sequence of points {zn*=rneiθn*} with limn→∞θn*=θ0*, such that η1:=δ((b-a)z,θ0*)>0. Indeed, we may suppose, without less of generality, that
(40)η∶=δ((b-a)z,θ)>0,θ∈(θ0+2kπ,θ0+(2k+1)π),η∶=δ((b-a)z,θ)<0,θ∈(θ0+(2k-1)π,θ0+2kπ)
with k∈ℤ. When n is large enough, we have |θn-θ0|≤l0, where l0 is a small constant. Choose now θn* such that l0/2≤θn*-θn≤l0. Then θ0+l0/2≤θ0*≤θ0+l0. For sufficiently large n, we have (12) for zn* and η1:=δ((b-a)z,θ0*)>0. Therefore
(41)|H(zn*)f(zn*)|≤M(rn,H)M(rn,f)1/5⟶0,(42)exp{1-ɛ2η1rn}≤|A0(zn*)A1(zn*)e(b-a)zn*|≤exp{3(1+ɛ)2η1rn}
for sufficiently large n. Taking now l0 small enough, we have δ(az,θ0*)>0, by the continuity of δ(az,θ). This yields
(43)exp{1-ɛ2δ(az,θ0*)rn}≤|A1(zn*)eazn*|≤exp{3(1+ɛ)2δ(az,θ0*)rn}.
Similarly as (36), a contradiction easily follows.
Case  2. Suppose now that δ(az,θ0)<0. Then, from the continuity of δ(az,θ) and (10), we have
(44)exp{3(1+ɛ)2δ(az,θ0)rn}≤|A1(zn)eazn|≤exp{1-ɛ2δ(az,θ0)rn}
for n large enough.
Subcase  2.1. Assume first that δ(bz,θ0)>0. From the continuity of δ(bz,θ) and (9), we deduce that
(45)exp{1-ɛ2δ(bz,θ0)rn}≤|A0(zn)ebzn|≤exp{3(1+ɛ)2δ(bz,θ0)rn}
for n large enough. From (4), we have
(46)|A0ebz|≤|f′′f|+|A1eaz||f′f|+|Hf|.
Substituting  (24), (28), (29), (44), and (45) into (46), we obtain
(47)exp{1-ɛ2δ(bz,θ0)rn}≤exp{2rnα0+ɛ1}rn-2(1+O(1))+exp{1-ɛ2δ(az,θ0)rn}×exp{rnα0+ɛ1}rn-1(1+O(1))+ɛ0.
Since α0+ɛ1<1, we see that (47) is contradictory as n→∞.
Subcase  2.2. Assume that δ(bz,θ0)<0. From the continuity of δ(bz,θ) and (9), we deduce that
(48)exp{3(1+ɛ)2δ(bz,θ0)rn}≤|A0(zn)ebzn|≤exp{1-ɛ2δ(bz,θ0)rn}
for n large enough. From (4), we have
(49)|f′′f|≤|A0ebz|+|A1eaz||f′f|+|Hf|.
Substituting  (24), (28), (29), (44), and (48) into (49), we obtain
(50)(νf(rn)rn)2(1+O(1))≤exp{1-ɛ2δ(bz,θ0)rn}+exp{1-ɛ2δ(az,θ0)rn}×exp{rnα0+ɛ1}rn-1(1+O(1))+ɛ0
as n→∞. Since α0+ɛ1<1, this implies that νf(r)→0, n→∞, which is impossible.
Subcase  2.3. Assume that δ(bz,θ0)=0. Arguing similarly as in Subcase  1.3, we may again construct another sequence of points {zn*=rneiθn*} with limn→∞θn*=θ0*, such that δ(az,θ0*)<0<δ(bz,θ0*). Replace δ(az,θ0) with δ(az,θ0*) in (44) and δ(bz,θ0) with δ(bz,θ0*) in (45), respectively. We obtain (44) and (45) for the sequence of {zn*=rneiθn*}. Similarly as (47), we get a contradiction as n→∞.
Case  3. In this final case, we suppose that δ(az,θ0)=0. We discuss three subcases according to δ(bz,θ0) as follows.
Subcase  3.1. Suppose that δ(bz,θ0)>0. By an argument similar to that in Subcase  1.3, we can choose another sequence of points {zn*=rneiθn*} with limn→∞θn*=θ0*, and l0/2≤θ0*-θ0≤l0, such that δ(az,θ0*)<0<δ(bz,θ0*). Similarly as in Subcase  2.3, a contradiction follows as n→∞.
Subcase  3.2. Suppose that δ(bz,θ0)<0. By an argument similar to the Subcase  3.2 of the proof of Theorem 1.1 in [7], we can choose another sequence of points {zn*=rneiθn*}, with limn→∞θn*=θ0*, such that δ(bz,θ0*)<0<δ(az,θ0*). From (4), for {zn*=rneiθn*}, we get
(51)|f′(zn*)f(zn*)|≤|1A1(zn*)eazn*|×(|A0(zn*)ebzn*|+|f′′(zn*)f(zn*)|+|H(zn*)f(zn*)|).
Replace δ(az,θ0) with δ(az,θ0*) in (32) and δ(bz,θ0) with δ(bz,θ0*) in (48), respectively. We obtain (32) and (48) for the sequence of {zn*=rneiθn*}. Substituting them into (51), this implies that νf(r)→0, n→∞, which is impossible.
Subcase  3.3. Finally, suppose that δ(bz,θ0)=0. We now have a/b=c∈ℝ, c≠0,1, and so az=cbz, (b-a)z=(1-c)bz.
Ifc<0, we may choose another sequence such that δ(bz,θ)<0<δ(az,θ). By an argument similar to that in Subcase 3.2, we can get νf(r)→0, n→∞, a contradiction.
If0<c<1, we similarly obtain δ((b-a)z,θ)>0 and δ(az,θ)>0 for another sequence. By an argument similar to that in Subcase  1.3, a contradiction follows.
Finally, ifc>1, we obtain δ((b-a)z,θ)<0<δ(az,θ) for another sequence. Similarly as in Subcase  1.2, a contradiction again follows.
Thus, we complete the proof when0<α0<1. When α0=0, we have (30). Similarly as the case when 0<α0<1, it results in contradiction. Hence, we get σ2(f)=α0=1.Proof of Theorem3.
Suppose thatf is a nontrivial solution of (5), and then f is an entire function. Since ϱ=max{σ(D0),σ(D1)}<1, we have
(52)|Dj(z)|≤exp{rϱ+ɛ}(j=0,1)
for any such that 0<3ɛ<1-ϱ. Similarly as in the proof of Theorem 1, we may choose a sequence of points {zn=rneiθn} that satisfy |f(zn)|=M(rn,f), with limn→∞θn=θ0, rn∈E∖(E1∪E2∪E3), rn→∞.
Step  1. We will prove that σ2(f)≤1. Since σ(A0;A1;H)<1, set σ(H)=λ<1. Then for any given ɛ satisfying ɛ<min{1-λ,(1-ϱ)/3}, when r is sufficiently large, we have
(53)|A1eaz+D1(z)|≤exp{r1+ɛ},|A0ebz+D0(z)|≤exp{r1+ɛ},(54)|H(z)|≤exp{rλ+ɛ}.
From the Wiman-Valiron theory, we have (24). By Theorem C, we know that σ(f)=∞. So we have (28). From (5) we have
(55)f′′f+(A1eaz+D1)f′f+(A0ebz+D0)=Hf.
Substituting  (24), (28), and (53) into (55), we obtain (26) and (27); thus, we have σ2(f)≤1.
Step  2. Set σ2(f)=α0, and we assert that α0=1. Now we assume that α0<1, and prove that σ2(f)=α0<1 results in contradiction. By Lemma 5, we have (29) and (30). Next we only prove the case 0<σ2(f)=α0≤1 by using (29). The case σ2(f)=α0=0 also can be proved by the same method, a little different is that we use (30) instead of (29). Since a=bc,c(<0) is a real number, there are three cases to be discussed, according to the signs of δ(az,θ0) and δ(bz,θ0).
Case  1. First assume that δ(bz,θ0)<0<δ(az,θ0), so we have (32) and (48). Combining (52), (32), and (48), we deduce
(56)|A0(zn)ebzn+D0(zn)|≤exp{rϱ+2ɛ},(57)|A1(zn)eazn+D1(zn)|≥exp{1-2ɛ2δ(az,θ0)rn}
provided thatn is large enough. From (5), we have
(58)|f′(zn)f(zn)|≤1|A1(zn)eazn+D1(zn)|×(|f′′(zn)f(zn)|+|A0(zn)ebzn+D0(zn)|+|H(zn)f(zn)|).
Substituting  (24), (28), (56), and (57) into (58), we obtain
(59)νf(rn)rn(1+O(1))≤exp{-1-2ɛ2δ(az,θ0)rn}×(exp{2rnα0+ɛ1}rn-2(1+O(1))+exp{rnϱ+2ɛ}+ɛ0)
for n large enough. Since α0+ɛ1<1, this implies that νf(rn)→0, n→∞, which is impossible.
Case  2. Next, assume that δ(az,θ0)<0<δ(bz,θ0), so we have (44) and (45). Combining (52), (44), and (45), we deduce
(60)|A0(zn)ebzn+D0(zn)|≥exp{1-2ɛ2δ(bz,θ0)rn},(61)|A1(zn)eazn+D1(zn)|≤exp{rnϱ+2ɛ}
forn is large enough. From (5), we have
(62)|A0(zn)ebzn+D0(zn)|≤|f′′(zn)f(zn)|+|A1(zn)eazn+D1(zn)|×|f′(zn)f(zn)|+|H(zn)f(zn)|.
Substituting  (24), (28), (60), and (61) into (62), we obtain
(63)exp{1-2ɛ2δ(bz,θ0)rn}≤exp{2rnα0+ɛ1}rn-2(1+O(1))+exp{rnϱ+2ɛ}exp{rnα0+ɛ1}rn-1(1+O(1))+ɛ0
for n large enough. Since α0+ɛ1<1, ϱ+2ɛ<1, this leads to a contradiction.
Case  3. Finally, we have to assume that δ(az,θ0)=δ(bz,θ0)=0. Similarly as in Subcase  1.3 of the proof of Theorem 1, we may again construct a sequence of points {zn*=rneiθn*}, with limn→∞θn*=θ0*, such that δ(az,θ0*)<0. Indeed, without loss of generality,
(64)δ(az,θ)>0,θ∈(θ0+2kπ,θ0+(2k+1)π),δ(az,θ)<0,θ∈(θ0+(2k-1)π,θ0+2kπ)
for allk∈ℤ. Provided that n is large enough, we have |θn-θ0|≤l0. Choosing now θ0* such that l0/2≤θn-θn*≤l0, then l0/2≤θ0-θ0*≤l0, thus, θ0-l0≤θ0*≤θ0-l0/2, and δ(az,θ0*)<0. Since now δ(bz,θ0*)>0, a contradiction follows as in Case  2 above.



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