# Nodal Solutions of thep-Laplacian with Sign-Changing Weight

Authors: Ruyun Ma; Xilan Liu; Jia Xu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We are concerned with determining values ofγ, for which there exist nodal solutions of the boundary value problem (|u′|p-2u′)′+γm(t)f(u)=0,t∈(0,1), u(0)=u(1)=0, where m∈C[0,1] is a sign-changing function, f:ℝ→ℝ with f(s)s>0. The proof of our main results is based upon global bifurcation techniques.


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## Body

## 1. Introduction

In [1], Ma and Thompson considered determining values of r, for which there exist nodal solutions of the boundary value problem
(1)u′′+rm(t)f(u)=0,t∈(0,1),u(0)=u(1)=0,
under the following assumptions:(H1)
f∈C(ℝ,ℝ) with sf(s)>0 for s≠0;(H~2)
m:[0,1]→[0,+∞) is continuous and does not vanish identically on any subinterval of [0,1];(H~3)
there existf0, f∞∈(0,+∞) such that
(2)f0=lim|s|→0f(s)s,f∞=lim|s|→+∞f(s)s.Using the bifurcation theory of Rabinowitz [2, 3], they proved the following.Theorem 1.
Let (H1), (H~2), and (H~3) hold. Assume that, for some k∈ℕ, either
(3)λkf∞<r<λkf0orλkf0<r<λkf∞.
Then (1) has two solutions uk+ and uk- such that uk+ has exactly k-1 zeros in (0,1) and is positive near 0 and uk- has exactly k-1 zeros in (0,1) and is negative near 0.The results of Theorem1 have been extended to the case that the weight function changes its sign by Ma and Han [4]. Bifurcation methods have been applied to study the existence of nodal solutions of nonlinear two-point, multipoint, and periodic boundary value problems; see [5–9] and the references therein. The results they obtained extend some well-known theorems of the existence of positive solutions for the related problems [10].However, no results on the existence of nodal solutions, even positive solutions, have been established for one-dimensionalp-Laplacian equation with sign-changing weight m(t). It is the purpose of this paper to establish a similar result to Theorem 1 for one-dimensional p-Laplacian equation with sign-changing weight. Problem with sign-changing weight arises from the selection-migration model in population genetics. In this model, m(t) changes sign corresponding to the fact that an allele A1 holds an advantage over a rival allele A2 at the same points and is at a disadvantage at others; the parameter r corresponds to the reciprocal of diffusion; for details see [11].Ifm(t)≡1, Del Pino et al. [12] established the global bifurcation theory for one-dimensional p-Laplacian eigenvalue problem. Peral [13] got the global bifurcation theory for p-Laplacian eigenvalue problem on the unite ball. In [14], Del Pino and Manásevich obtained the global bifurcation from the principal eigenvalue for p-Laplacian eigenvalue problem on the general domain. If m(t)≥0 and is singular at t=0 or t=1, Lee and Sim [15] also established the bifurcation theory for one-dimensional p-Laplacian eigenvalue problem. However, if m(t) changes sign, there are a few papers dealing with the p-Laplacian eigenvalue problem via bifurcation techniques. In [16], Drábek and Huang established the global bifurcation from the principal eigenvalue for p-Laplacian eigenvalue problem in ℝN.The purpose of this paper is to study the bifurcation behavior of one-dimensionalp-Laplacian eigenvalue problem as follows:
(4)φp(u′)′+γm(t)f(u)=0,t∈(0,1),u(0)=u(1)=0,
under the condition (H1) and(H2)
m(t)∈C[0,1] changes sign and
(5)meas{x∈[0,1]∣m(t)=0}=0;(H3)
there existsf0∈(0,∞) such that
(6)f0=lim|s|→0f(s)φp(s),where φp(s)=|s|p-2s with 1<p<+∞;(H4)
there existsf∞∈(0,+∞) such that
(7)f∞=lim|s|→+∞f(s)φp(s).Moreover, based on our global bifurcation theorem, we will prove the existence of nodal solutions for the corresponding nonlinear problem with a parameter (see Theorem 11).The main tool is the global bifurcation techniques in [17].The rest of this paper is arranged as follows. In Section2, we establish the global bifurcation theory for one-dimensional p-Laplacian eigenvalue problem with sign-changing weight. In Section 3, we state and prove the main results of this paper.


## 2. Some Preliminaries

LetE be the Banach space C01[0,1] with the norm
(8)∥u∥=max{∥u∥∞,∥u′∥∞}.LetY=L1(0,1) with its usual normal ∥·∥L1.We start by considering the following auxiliary problem:(9)φp(u′)′=h,t∈(0,1),u(0)=u(1)=0,
for a given h∈L1(0,1). By a solution of problem (9), we understand a function u∈E with φp(u′) absolutely continuous which satisfies (9). Problem (9) is equivalently written to
(10)u(t)=Gp(h)(t)∶=∫0t‍φp-1(a(h)+∫0s‍h(τ)dτ)ds,
where a:Y→ℝ is a continuous function satisfying
(11)∫01‍φp-1(a(h)+∫0s‍h(τ)dτ)ds=0.
It is known that Gp:Y→E is continuous and maps equi-integrable sets of Y into relatively compacts of E. One may refer to Lee and Sim [15] for details.Since the bifurcation points of(12)φp(u′(t))′+λm(t)f(u(t))=0a.e.in(0,1),u(0)=u(1)=0
is related to the eigenvalues of the problem
(13)φp(u′(t))′+λm(t)φp(u(t))=0a.e.in(0,1),u(0)=u(1)=0.We define the operatorTλp:E→E by
(14)Tλp(u)(t)=∫0t‍φp-1(∫0s‍a(-λmφp(u(τ)))Tλp(u)n(t)=∫0t‍φp-1-∫0s‍λm(τ)φp(u(τ))dτ)dsTλp(u)(t)=:Gp(-λmφp(u))(t).
Then Tλp:E→E is completely continuous and problem (13) is equivalent to
(15)u=Tλp(u).The following spectrum result plays a fundamental role in our study.Lemma 2 (see [18, 19]).
Let (H2) hold. Then(i)
the set of all eigenvalues of the problem (13) is two infinite sequences of simple eigenvalues as follows:
(16)0<μ1+(p)<μ2+(p)<⋯<μk+(p)<⋯,limk→+∞μk+(p)=+∞,0>μ1-(p)>μ2-(p)>⋯>μk-(p)>⋯,limk→+∞μk-(p)=-∞;(ii)
fork∈ℕ and ν∈{+,-}, Ker(I-Tμkν(p)p) is a space of E with dimensional 1;(iii)
the eigenfunction corresponding toμkν(p) has exactly k-1 simple zeros in (0,1).Remark 3.
Using the Gronwall inequality, we can easily show that all zeros of eigenfunction corresponding to eigenvalueμkν(p) are simple.
It is very known thatTλ2 is completely continuous in C1[0,1]. Thus, the Leray-Schauder degree dLS(I-Tλ2,Br(0),0) is well-defined for arbitrary r-ball Br(0) and λ≠μkν,k∈ℤ and ν∈{+,-}.Lemma 4.
Forr>0, we have
(17)dLS(I-Tλ2,Br(0),0)={1,ifλ∈(μ1-(2),μ1+(2)),(-1)k,ifλ∈(μk+(2),μk+1+(2)),k∈ℕ,(-1)k,ifλ∈(μk+1-(2),μk-(2)),k∈ℕ.Proof.
We divide the proof into two cases.
Case  1. λ≥0. Since Tλ2 is compact and linear, by [20, Theorem 8.10] and Lemma 2 (ii) with p=2,
(18)dLS(I-Tλ2,Br(0),0)=(-1)m(λ),
where m(λ) is the sum of algebraic multiplicity of the eigenvalues μ of (13) satisfying μ-1λ>1.
Ifλ∈[0,μ1+(2)), then there are no such μ at all; then
(19)dLS(I-Tλ2,Br(0),0)=(-1)m(λ)=(-1)0=1.
Ifλ∈(μk+(2),μk+1+(2)) for some k∈ℕ, then
(20)(μj+(2))-1λ>1,j∈{1,⋯,k}.
This together with Lemma2 (ii) implies the following:
(21)dLS(I-Tλ2,Br(0),0)=(-1)k.
Case  2. λ<0. In this case, we consider a new sign-changing eigenvalue problem as follows
(22)u′′+λ^m^(t)u=0,t∈(0,1),u(0)=u(1)=0,
where λ=-λ, m^(t)=-m(t). It is easy to check that
(23)μ^k+(2)=-μk-(2),k∈ℕ.
Thus, we may use the result obtained in Case 1 to deduce the desired result.We first show that the principle eigenvalue functionμ1ν:(1,+∞)→ℝ is continuous.Proposition 5.
The eigenvalue functionμ1ν:(1,+∞)→ℝ is continuous.Proof.
We only show thatμ1+:(1,+∞)→ℝ is continuous since the case of μ1- is similar. In the following proof, we will shorten μ1+ to μ1. From the variational characterization of μ1(p), it follows that
(24)μ1(p)=sup{μ>0∣μ∫01‍m(t)|u|pdtmmmm≤∫01‍|u′|pdt,∀u∈Cc∞(0,1)}.
Let{pj}j=1∞ be a sequence in (1,+∞) convergent to p>1. We will show that
(25)limj→+∞μ1(pj)=μ1(p).
To do this, letu∈Cc∞(0,1). Then, from (24),
(26)μ1(pj)∫01‍m(t)|u|pjdt≤∫01‍|u′|pjdt.
On applying the Dominated Convergence Theorem, we find that(27)limsupj→+∞μ1(pj)∫01‍m(t)|u|pdt≤∫01‍|u′|pdt.
Relation (27), the fact that u is arbitrary and (24) yield
(28)limsupj→+∞μ1(pj)≤μ1(p).
Thus, to prove (25), it suffices to show that
(29)liminfj→+∞μ1(pj)≥μ1(p).
Let {pk}k=1∞ be a subsequence of {pj}j=1∞ such that limk→+∞μ1(pk)=liminfj→+∞μ1(pj).
Let us fixɛ0>0 so that p-ɛ0>1 and, for each 0<ɛ<ɛ0, W01,p-ɛ(0,1) is compactly embedded into Lp+ɛ(0,1). For k∈ℕ, let us choose uk∈W01,pk(0,1) such that
(30)∫01‍|uk′|pkdt=1,(31)∫01‍|uk′|pkdt=μ1(pk)∫01‍m(t)|uk|pkdt.
For 0<ɛ<ɛ0, there exists k0∈ℕ such that p-ɛ<pk<p+ɛ for any k≥k0. Thus, for k≥k0, (30) and Hölder's inequality imply that
(32)∫01‍|uk′|p-ɛdt≤1.
This shows that {uk}k=k0∞ is a bounded sequence in W01,p-ɛ(0,1). Passing to a subsequence if necessary, we can assume that uk⇀u in W01,p-ɛ(0,1) and hence that uk→u in Lp+ɛ(0,1). Furthermore, u∈Lp(0,1) and uk→u in Lpk(0,1) for k≥k0. It follows that
(33)|∫01|uk|pkdt-∫01‍|u|pkdt|≤∫01pk|u+θuk|pk-1|uk-u|dt≤(p+ɛ)(∫01|u+θuk|pkdt)(pk-1)/pk×(∫01|uk-u|pkdt)1/pk≤(p+ɛ)(∥u∥pk+∥uk∥pk)pk-1(∫01|uk-u|pkdt)1/pk⟶0
as k→+∞. It is clear that
(34)∫01|u|pkdt-∫01|u|pdt⟶0ask⟶+∞.
Thus,(35)∫01|uk|pkdt⟶∫01|u|pdt.
Similarly, we can also obtain that(36)∫01‍m+(t)|uk|pkdt⟶∫01‍m+(t)|u|pdt,∫01‍m-(t)|uk|pkdt⟶∫01‍m-(t)|u|pdt,
where m+(t)=max{m(t),0} and m-(t)=-min{m(t),0}. Therefore,
(37)∫01m(t)|uk|pkdt=∫01m+(t)|uk|pkdt-∫01m-(t)|uk|pkdt⟶∫01m+(t)|u|pdt-∫01m-(t)|u|pdt=∫01m(t)|u|pdt.
We note that (30) and (31) imply that
(38)μ1(pk)∫01‍m(t)|uk|pkdt=1
for all k∈ℕ. Thus, letting k go to +∞ in (38) and using (37), we find that
(39)liminfj→+∞μ1(pk)∫01‍m(t)|u|pdt=1.
On the other hand, sinceuk⇀u in W01,p-ɛ(0,1), from (32) we obtain that
(40)∥u′∥p-ɛp-ɛ≤liminfk→+∞∥uk′∥p-ɛp-ɛ≤1ɛ/p.
Now, lettingɛ→0+ and applying Fatou's Lemma, we find that
(41)∥u′∥pp≤1.
Hence, u∈W1,p(0,1); here W1,p(0,1) denotes the radially symmetric subspace of W1,p(0,1). We claim that actually u∈W01,p(0,1). Indeed, we know that u∈W01,p-ɛ(0,1) for each 0<ɛ<ɛ0. For ϕ∈Cc∞(ℝ), it is easy to see that
(42)|∫01‍uϕ′dt|≤∥u′∥p-ɛ∥ϕ∥(p-ɛ)′,i=1,…,N.
Then, letting ɛ→0+, we obtain that
(43)|∫01‍uϕ′dt|≤∥u′∥p∥ϕ∥p′,i=1,…,N,
where p′=p/(p-1). Since ϕ is arbitrary, from Proposition IX-18 of [21], we find that u∈W01,p(0,1), as desired.
Finally, combining (39) and (41), we obtain that
(44)liminfj→+∞μ1(pk)∫01‍m(t)|u|pdt≥∫01‍|u′|pdt.
This and the variational characterization ofμ1(p) imply (29) and hence (25). This concludes the proof of the lemma.Using Remark3, Lemma 2, and Proposition 5, we will show that all eigenvalue functions μk±:(1,+∞)→ℝ, 2≤k∈ℕ are continuous.Lemma 6.
For fixed2≤k∈ℕ and ν∈{+,-}, μkν(p) as a function of p∈(1,+∞) is continuous.Proof.
Letukν be an eigenfunction corresponding to μkν(p). By Lemma 2 and Remark 3, we know that u has exactly k-1 simple zeros in I; that is, there exist ck,1,…,ck,k-1∈I such that u(ck,1)=⋯=u(ck,k-1)=0. For convenience, we set ck,0=0, ck,k=1, and Ji=(ck,i-1,ck,i) for i=1,…,k. Let μ1ν(p,m/Ji,Ji) denote the first positive or negative eigenvalue of the restriction of problem (13) on Ji for i=1,…,k. Lemma 3 of [18] follows that μkν(p)=μ1ν(p,m/Ji,Ji) for i=1,…,k. Using a similar proof to Proposition 5, we can show that μ1ν(p,m/Ji,Ji) is continuous with respect to p for i=1,…,k. Therefore, μkν(p) is also continuous with respect to p.Lemma 7.
(i) Let{μk+(p)}k∈ℕ be the sequence of positive eigenvalues of (13). Let λ be a constant with λ≠μk+(p) for all k∈ℕ. Then, for arbitrary r>0,
(45)deg(Tλp,Br(0),0)=(-1)β,
where β is the number of eigenvalues μn+(p) of problem (13) less than λ.
(ii) Let{μk-(p)}k∈ℕ be the sequence of negative eigenvalues of (13). Consider λ≠μk-(p), k∈ℕ; then
(46)deg(Tλp,Br(0),0)=(-1)β,∀r>0,
where β is the number of eigenvalues μk-(p) of problem (25) larger than λ.Proof.
We will only prove the caseλ>μ1+(p) since the proof for the other cases is similar. We also only give the proof for the case p>2. Proof for the case 1<p<2 is similar. Assume that μk+(p)<λ<μk+1+(p) for some k∈ℕ. Since the eigenvalues depend continuously on p, there exists a continuous function χ:[2,p]→ℝ and q∈[2,p] such that μk+(q)<χ(q)<μk+1+(q) and λ=χ(p). Define
(47)Φ(q,u)=u-Gq(-χ(q)m(t)φq(u)).
It is easy to show thatΦ(q,u) is a compact perturbation of the identity such that, for all u≠0, by definition of χ(q), Φ(q,u)≠0, for all q∈[2,p]. Hence, the invariance of the degree under homotopology and the classical result for p=2 imply
(48)deg(Tλp,Br(0),0)=deg(Tλ2,Br(0),0)=(-1)k.For the existence of bifurcation branches for (12), we will make use of the following global bifurcation theorem results.Lemma 8 (see [17]).
LetX be a Banach space. Let F:ℝ×X→X be completely continuous such that F(λ,0)=0 for all λ∈ℝ. Suppose that there exist constants ρ, η∈ℝ, with ρ<η, such that (ρ,0) and (η,0) are not bifurcation points for the equation
(49)u-F(λ,u)=0.
Furthermore, assume that
(50)deg(I-F(ρ,·),Br(0),0)≠deg(I-F(η,·),Br(0),0),
where Br(0)={u∈X:∥u∥<r} is an isolating neighborhood of the trivial solution for both constants ρ and η. Let
(51)𝒮={(λ,u):(λ,u)isasolutionof(49)withu≠0}¯𝒮∪([ρ,η]×{0}),
and let 𝒞 be the component of 𝒮 containing [ρ,η]×{0}. Then, either (i)
𝒞 is unbounded in ℝ×X or(ii)
𝒞∩[(ℝ∖[ρ,η])×{0}]≠∅.Define the Nemytskii operatorsH:ℝ×E→Y by
(52)H(λ,u)(t)∶=-λm(t)f(u(t)).Then, it is clear thatH is continuous operator which sends bounded sets of ℝ×E into an equi-integrable sets of Y and problem (12) can be equivalently written as
(53)u=Gp∘H(λ,u)∶=F(λ,u).F is completely continuous in ℝ×E→E and F(λ,0)=0, for all λ∈ℝ.Notice that (12) with λ=0 has only the trivial solution. Applying this fact and Lemma 8 and the same method to prove [15, Theorem 2.1] with obvious changes, we may obtain the following.Lemma 9.
Assume that (H1), (H2), and (H3) hold. Then, for fixed p>1 and for fixed σ∈{+,-}, each (μkν(p)/f0,0) is a bifurcation point of (12) and the associated bifurcation branch (𝒞kν)σ satisfies the following; (1)
(𝒞kν)σ is unbounded in E;(2)
(𝒞kν)σ⊂(ℝ×Φkσ)∪{(μkν(p),0)}, where Φkσ is the set of function u∈C01[0,1] which has exact k-1 simple zeros in (0,1), and σu is positive near 0.Finally, we give a key lemma that will be used in Section3. Let
(54)I+:={t∈[0,1]∣m(t)>0},I-:={t∈[0,1]∣m(t)<0}.Lemma 10.
Let(H2) hold. Let I=[a,b] be such that I⊂I+ and
(55)measI>0.
Letgn:[0,1]→(0,+∞) be such that
(56)limn→+∞gn(t)=+∞,uniformlyonI.
Letyn∈E be a solution of the equation
(57)φp(yn′)′+m(t)gn(t)φp(yn)=0,t∈(0,1).
Then, the number of zeros of yn|I goes to infinity as n→+∞.Proof.
After taking a subsequence if necessary, we may assume that(58)m(t)gnj(t)≥j,t∈I,
as j→+∞. It is easy to check that the distance between any two consecutive zeros of any nontrivial solution of the equation
(59)φp(u′(t))′+jφp(u(t))=0,t∈I,
goes to zero as j→+∞. Using this with [21, Lemma 2.5], it follows the desired results.


## 3. Main Results and Its Proof

Letμk± be the kth positive or negative eigenvalue of (13). By applying Lemma 9, we will establish the main results as follows.Theorem 11.
Let (H1), (H2), (H3), and (H4) hold. Assume that, for some k∈ℕ, either
(60)γ∈(μk+(p)f∞,μk+f0)∪(μk-(p)f0,μk-(p)f∞)
or
(61)γ∈(μk+(p)f0,μk+(p)f∞)∪(μk-(p)f∞,μk-(p)f0).
Then, (4) has two solutions uk+ and uk- such that uk+ has exactly k-1 zeros in (0,1) and is positive near 0 and uk- has exactly k-1 zeros in (0,1) and is negative near 0.Proof.
We only prove the case ofγ>0. The case of γ<0 is similar. Consider the problem
(62)φp(u′)′+λγm(t)f(u)=0,t∈(0,1),u(0)=0,u(1)=0.
Considering the results of Lemma9, we have that, for each integer k≥1, σ∈{+,-}, there exists a continuum (Ck+)σ⊆Φkσ of solutions of (62) joining (μk+(p)/γf0,0) to infinity in (0,∞)×Φkσ. Moreover, (Ck+)σ∖{(μk+(p)/γf0,0)}⊂(0,∞)×Φkσ.
It is clear that any solution of (62) of the form (1,u) yields a solution u of (4). We will show that (Ck+)σ crosses the hyperplane {1}×E in ℝ×E. To this end, it will be enough to show that (Ck+)σ joins (μk+(p)/γf0,0) to (μk+(p)/γf∞,+∞). Let (ηn,yn)∈(Ck+)σ satisfy
(63)μn+∥yn∥⟶+∞.
We note that ηn>0 for all n∈ℕ since (0,0) is the only solution of (62) for λ=0 and (Ck+)σ∩({0}×E)=∅.
Case  1. μk+(p)/f∞<γ<μk+(p)/f0. In this case, we only need to show that
(64)(μk+(p)γf∞,μk+(p)γf0)⊆{μ∈ℝ:(μ,u)∈(Ck+)σ}.
We divide the proof into two steps.
Step  1. We show that, if there exists a constant number M>0 such that
(65)ηn⊂(0,M]
for n∈ℕ large enough, then (Ck+)σ joins (μk+(p)/γf0,0) to (μk+(p)/γf∞,+∞).
In this case, it follows that(66)∥yn∥⟶+∞.
Letξ∈C(ℝ) be such that
(67)f(u)=f∞φp(u)+ξ(u).
Then,
(68)lim|u|→+∞ξ(u)φp(u)=0.
Let
(69)ξ~(u)=max0≤|s|≤u|ξ(s)|.
Then, ξ~ is nondecreasing and
(70)limu→+∞ξ~(u)|u|p-1=0.
We divide the equation(71)φp(yn′)′-μnγm(t)f∞φp(yn)=μnγm(t)ξ(yn)
by ∥yn∥ and set y¯n=yn/∥yn∥. Since y¯n is bounded in E, after taking a subsequence if necessary, we have y¯n⇀y¯ for some y¯∈E and y¯n→y¯ in Y with ∥y¯∥=1. Moreover, from (70) and the fact that ξ~ is nondecreasing, we have
(72)limn→+∞ξ(yn(t))∥yn∥p-1=0,
since
(73)ξ(yn(t))∥yn∥p-1≤ξ~(|yn(t)|)∥yn∥p-1≤ξ~(∥yn(t)∥∞)∥yn∥p-1≤ξ~(∥yn(t)∥)∥yn∥p-1.
By the continuity and compactness ofGp, it follows that
(74)y¯=Gp(μ¯γm(t)f∞φp(y¯)),
where μ¯=limn→+∞μn, again choosing a subsequence and relabeling if necessary.
We claim that(75)y¯∈(Ck+)σ.
Suppose on the contrary thaty¯∈(Ck+)σ. Since y¯≠0 is a solution of (74) and all zeros of y¯ in [0,1] are simple, it follows that y¯∈(Ch+)ι≠(Ck+)σ for some h∈ℕ and ι∈{+,-}.
By the openness ofE∖(Ck+)σ, we have that there exists a neighborhood U(y¯,ρ0) such that
(76)U(y¯,ρ0)⊂E∖(Ck+)σ,
which contradicts the facts that y¯n→y¯ in E and y¯n∈(Ck+)σ. Therefore, y¯∈Ckν. Moreover, by Lemma 2, μ¯γf∞=μk+(p), so that
(77)μ¯=λkγf∞.
Therefore, (Ck+)σ joins  (μk+(p)/γf0,0) to (μk+(p)/γf∞,+∞).
Step  2. We show that there exists a constant M such that μn∈(0,M] for n∈ℕ large enough.
On the contrary, we suppose that(78)limn→+∞μn=+∞.
Since(ηn,yn)∈(Ck+)σ, it follows that
(79)φ(yn′)′+γηnm(t)f(yn)φ(yn)φ(yn)=0.
Let(80)0=τ(0,n)<τ(1,n)<⋯<τ(k,n)=1
be the zeros of yn in [0,1]. Then, after taking a subsequence if necessary,
(81)limn→+∞τ(l,n)∶=τ(l,∞),l∈{0,1,⋯,k-1}.
Notice that Lemma 10 and the fact that yn has exactly k-1 simple zeros in [0,1] yield
(82)[∪l=0k-1(τ(l,∞),τ(l+1,∞))]∩I+=∅,
which implies that
(83)meas{[∪l=0k-1(τ(l,∞),τ(l+1,∞))]∩I-}=1.
However, this contradicts (H2): 0<measI-<1.
Case  2. μk+(p)/f0<γ<μk+(p)/f∞. In this case, we have that
(84)μk+(p)γf0<1<μk+(p)γf∞.
Assume that(ηn,yn)∈(Ck+)σ is such that
(85)limn→+∞(μn+∥yn∥)=+∞.
Ifηn→+∞, then we are done!
If there existsM>0, such that, for n∈ℕ sufficiently large,
(86)ηn∈(0,M].
Applying the same method used in Step 1 of Case 1, after taking a subsequence and relabeling if necessary, it follows that(87)(ηn,yn)⟶(μk+(p)γf∞,+∞)asn⟶+∞.
Thus,(Ck+)σ joins (μk+(p)/γf0,0) to (μk+(p)/γf∞,+∞).



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*Source: 406350-2013-11-27.xml*