# Existence of Sign-Changing Solutions to Equations Involving the One-Dimensionalp-Laplacian

Authors: Ruyun Ma; Lingfang Jiang
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

We consider the equations involving the one-dimensionalp-Laplacian (P):  (u′tp-2u′(t))′+λf(u(t))=0, 0<t<1, and
u(0)=u(1)=0, where p>1,λ>0,f∈C1(R;R),f(s)s>0, and
s≠0. We show the existence of sign-changing solutions under the assumptions f∞=lim|s|→∞⁡(fs/sp-1)=+∞ and f0=lim|s|→0(f(s)/sp-1)∈[0,∞]. We also show that (P) has exactly one solution having specified nodal properties for λ∈(0,λ*) for some λ*∈(0,∞). Our main results are based on quadrature method.


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## Body

## 1. Introduction and Main Results

Existence and multiplicity of positive solutions of nonlinear second order boundary value problem(1)u′′(t)+λf(u(t))=0,0<t<1,u(0)=u(1)=0
and its generalized forms have been extensively studied via the fixed point theorem in cones, bifurcation theory, quadrature method, and fixed index theorem in the past four decades; see Erbe and Wang [1], Henderson and Wang [2], Laetsch [3], Fink et al. [4], Ma and Thompson [5, 6], and the references therein.Existence and multiplicity of positive solutions of the corresponding one-dimensionalp-Laplacian
(2)(|u′(t)|p-2u′(t))′+λa(t)f(u(t))=0,0<t<1,(3)u(0)=u(1)=0
have also been studied by several authors; see Lee and Sim [7], Wang [8], Kong and Wang [9], Aranda and Godoy [10], Bouguima and Lakmeche [11], and de Coster [12] for references along this line.Recently, Lee and Sim [7] consider the existence and multiplicity of positive solutions of (2), (3) under the assumptions
(4)f∞=∞,f0∈(0,∞).
They proved the following.Theorem A (see [7, Theorem 3.14]).
Assume (4) hold. Then, there exist λ*≥λ*>0 such that (2), (3) have at least one positive solution for λ<λ* and no positive solution for λ>λ*.Of course, natural question is as follows. What would happen if we allow thatf0∈{0,∞}?It is the purpose of this paper to study sign-changing solutions of (2), (3) under the assumptions a(t)≡1 and
(A)f∞=∞,f0=0,
or
(B)f∞=∞,f0=∞.
The main tool is the quadrature method.We will make the following assumptions:(H0)
f(s)s>0 for s≠0;(H1)
lim⁡|s|→∞(f(s)/sp-1)=+∞;(H2)
lim⁡s→+∞(((p-1)/p)f(s)-(s/p)f′(s))<0 and lim⁡s→-∞(((p-1)/p)f(s)-(s/p)f′(s))>0.Letλk:=kpπpp,k∈N, where πp:=2∫0(p-1)1/p‍1/(1-(sp/(p-1)))1/pds. The main results of this paper are the following.Theorem 1.
Let (H0), (H1), and (H2) hold. Assume thatf satisfies
(5)f0=lim⁡|s|→0f(s)sp-1∈(0,∞).
Then, for k∈N, (2), (3) have two solutions uk+ and uk- for each λ∈(0,λk/f0): uk+ has k-1 zeros in (0,1) and is positive near 0, and uk- has k-1 zeros in (0,1) and is negative near 0. Moreover, there exists a constant λ*∈(0,λk/f0), such that for each λ∈(0,λ*) the above solution is unique.Theorem 2.
Let (H0), (H1), and (H2) hold. Assume thatf0=0. Then, for k∈N, (2), (3) have two solutions uk+ and uk- for each λ∈(0,∞): uk+ has k-1 zeros in (0,1) and is positive near 0, and uk- has k-1 zeros in (0,1) and is negative near 0. Further, there exists a constant λ*∈(0,∞) independent of k, such that for each λ∈(0,λ*) the above solution is unique.Theorem 3.
Let (H0) and (H1) hold. Assume thatf0=∞. Then, for k∈N, there exists a constant λ* small independent of k, such that for each λ∈(0,λ*) (2), (3) have two solutions uk+ and vk+: uk+ and vk+ have k-1 zeros in (0,1) and are positive near 0; and problems (2), (3) have two solutions uk- and vk-, where uk- and vk- have k-1 zeros in (0,1) and are negative near 0.Remark 4.
Forp=2, the existence of positive and sign-changing solutions has been extensively studied by many authors [1–6], but they did not give any information about the uniqueness of nodal solutions.Remark 5.
It is worth noticing that Lee and Sim [7] studied the nonautonomous cases (2), (3) and obtained the existence of positive solutions with f∞=∞,f0∈(0,∞). They gave no information about the sign-changing solutions. In Theorem 1, we show the existence of solutions having specified nodal properties.Remark 6.
Very little is known in the available literature even in the special casep=2. We establish uniqueness results in this paper; see Theorems 1 and 2.Remark 7.
Let us consider the problem(6)(|u′(t)|8/3-2u′(t))′+λg(u(t))=0,0<t<1,u(0)=u(1)=0,
where g(s)=s11/3+s5/3. Obviously, g satisfies (H0) and (H1). Since
(7)8/3-18/3g(s)-s8/3g′(s)=-34s11/3,
it is easy to see that (H2) is fulfilled. Thus, Theorem 1 implies that, for k∈N, (6) have two solutions uk+ and uk- for each λ∈(0,λk/f0): uk+ has k-1 zeros in (0,1) and is positive near 0, and uk- has k-1 zeros in (0,1) and is negative near 0. Moreover, there exists a constant λ*∈(0,λk/f0), such that for each λ∈(0,λ*) the above solution is unique.For other results dealing withp-Laplacian operators and the bifurcation behavior of solutions, see [13–24] and the references therein.The rest of the paper is arranged as follows. In Section2, we state and prove some preliminary results. Finally, in Section 3, we give the proofs of Theorems 1, 2, and 3.


## 2. Quadrature Method and Preliminaries

Letf∈C1(R;R),f(s)s>0 for s≠0 and F(s)=∫0s‍f(t)dt.Lemma 8.
Ifu is any solution of (2), (3) and x0∈(0,1) is such that u′(x0)=0, then u(x0-t)=u(x0+t),t∈[0,min⁡{x0,1-x0}].Proof.
Sincef is autonomous, both u(x0-t) and u(x0+t) satisfy the initial value problem
(8)(|w′(t)|p-2w′(t))′+λf(w(t))=0,iiiiiiiiiiiiiiit∈[0,min⁡{x0,1-x0}],w(0)=u(x0),w′(0)=0.
By Reichel and Walter [14, Theorem 2] and [14, (iii) and (v) in the case (β) of Theorem 4], (8) has a unique solution defined on t∈(0,min⁡{x0,1-x0}). Therefore, u(x0-t)=u(x0+t).Now, we divide the discussion into two cases.Case 1 (k=2n+1.).
In this case, we attempt to find a solution of (2), (3) with 2n zeros in (0, 1) and u′(0)<0 and a solution of (2), (3) with 2n zeros in (0, 1) and u′(0)>0.
Obviously, ifu is a sign-changing solution with 2n zeros in (0,1) and u′(0)<0, then, thanks to Lemma 8 and the fact that (2) is autonomous, we only need to study u on the intervals [x0,2x0] and [2x0,1/2n+((n-1)/n)x0].Multiplying (2) throughout by u′(t), we obtain
(9)(|u′|p-2u′(t))′u′(t)+λf(u)u′(t)=0,
and integrating we have
(10)|u′|p=-λpp-1F(u(t))+pp-1c.
If h=sup⁡t∈[0,1]u(t) and q=inf⁡t∈[0,1]u(t), then u(x0)=q and u(1/2n+((n-1)/n)x0)=h. Substituting t=x0 and t=1/2n+((n-1)/n)x0 in (10), we get c=λF(q) and c=λF(h). Hence, q=q(h) is such that
(11)F(q)=F(h).
Thus, we have
(12)u′(t)=(pp-1λ)1/p(F(h)-F(u))1/p;iiiiiiiiiiiiiiiiit∈(2x0,12n+n-1nx0),(13)u′(t)=(pp-1λ)1/p(F(q)-F(u))1/p;t∈(x0,2x0).
Integrating (12) and (13) on (2x0,1/2n+((n-1)/n)x0) and (x0,2x0), respectively, we obtain
(14)∫0u(t)‍du(F(h)-F(u))1/p=(pp-1λ)1/p(t-2x0),(15)∫u(t)0‍du(F(q)-F(u))1/p=(pp-1λ)1/p(2x0-t).
Hence, substituting t=1/2n+((n-1)/n)x0 in (14) and t=x0 in (15), we have
(16)∫0h‍du(F(h)-F(u))1/p=(pp-1λ)1/p(12n-n+1nx0),(17)∫q0‍du(F(q)-F(u))1/p=(pp-1λ)1/px0.
Multiplying (17) by (n+1)/nand adding to (16), we can see that λ and h satisfy
(18)(λ)1/p=2(p-1p)1/p{n∫0h‍du(F(h)-F(u))1/piiiiiiiiiiiiiiiiiiiiiiiii+(n+1)∫q0‍du(F(q)-F(u))1/p}:=G2n(h).
In fact, the following result holds.Lemma 9.
Givenλ>0, if there exists h∈(0,∞) such that G2n(h)=(λ)1/p, then (2), (3) have a sign-changing solution with 2n interior zeros satisfying ∥u∥=sup⁡t∈(0,1)u(t)=h. Further, G2n(h) is a continuous function in (0,∞) and it is also differentiable with the derivative given by
(19)dG2n(h)dh=2(p-1p)1/p×{n∫01‍H(h)-H(hv)(F(h)-F(hv))(p+1)/pdviiiiiiii-(n+1)dqdh∫01‍H(q)-H(qv)(F(q)-F(qv))(p+1)/pdv},
where H(s)=F(s)-(s/p)f(s).The proof of the above theorem follows by carefully extending the arguments used in [15, Theorem 2.2] for second order differential equation to the case of one-dimensional p-Laplacian.Using the same argument, with obvious changes, we may deduce the following.Ifu is a sign-changing solution with 2n zeros in (0,1) and u′(0)>0, the corresponding G12n(h) is
(20)G12n(h)=2(p-1p)1/p×{(n+1)∫0h‍du(F(h)-F(u))1/piiiiiiii+n∫q0‍du(F(q)-F(u))1/p}.Case 2 (k=2n).
In this case, ifu is a sign-changing solution with 2n-1 zeros in (0,1) and u′(0)>0, the corresponding G2n-1(h) is
(21)G2n-1(h)=(λ)1/p=2(p-1p)1/p×{n∫0h‍du(F(h)-F(u))1/piiiiiiiiiiiiiiiiiiiiiiiiii+n∫q0‍du(F(q)-F(u))1/p}.
Similarly, we may get the same function G2n-1(h) as above when u is a sign-changing solution with (2n-1) zeros in (0,1) with u′(0)<0.


## 3. The Proofs of the Main Results

Proof of Theorem1.
First, we considerk=2n+1.
It follows from the quadrature method that a solution with2n zeros in (0,1) exists if for λ>0 there exists h∈(0,∞) such that (λ)1/p=G2n(h). To prove this, we will show that (0,(2n+1)πp/f01/p)⊂Range(G2n(h)). We achieve this by proving(A)
lim⁡h→+∞G2n(h)=0,(B)
lim⁡h→0G2n(h)=(2n+1)πp/f01/p.
Proof of (A). Recall that
(22)G2n(h)=2(p-1p)1/p×{n∫0h‍du(F(h)-F(u))1/piiiiiii+(n+1)∫q0‍du(F(q)-F(u))1/p}.
First let us consider(23)2(p-1p)1/pn∫0h‍du(F(h)-F(u))1/p=2(p-1p)1/pn∫0h‍du(∫uh‍f(t)dt)1/p.
To this end, we have from (H1) that, for anyk∈N, there exists Rk∈(0,∞), such that
(24)f(s)≥kpsp-1,s≥Rk.
Ifh>Rk, it follows from (23) and (24) that we have that
(25)2(p-1p)1/p∫0Rk‍ds(F(h)-F(s))1/p+2(p-1p)1/pn∫Rkh‍ds(F(h)-F(s))1/p=2(p-1p)1/pn∫0Rk‍ds(∫sh‍f(w)dw)1/p+2(p-1p)1/pn∫Rkh‍ds(∫sh‍f(w)dw)1/p≤2(p-1p)1/pn∫0Rk‍(kpp(hp-Rkp))-1/pds+2(p-1p)1/pn∫Rkh‍(kpp(hp-sp))-1/pds=2(p-1)1/pn1k×{Rk[(hp-Rkp)]-1/p+∫Rk/h1‍1(1-vp)1/pdv}=n1k{1(1-(vp/(p-1)))1/p2(p-1)1/pRk[(hp-Rkp)]-1/piiiiiiiiiiiiiii+2∫Rk(p-1)1/p/h(p-1)1/p‍1(1-(vp/(p-1)))1/pdv}⟶nCkash⟶∞,
where
(26)C≔lim⁡h→∞{1(1-(vp/(p-1)))1/p2(p-1)1/pRk[(hp-Rkp)]-1/piiiiiiiiiiiiii+2∫Rk(p-1)1/p/h(p-1)1/p‍1(1-(vp/(p-1)))1/pdv}=πp.
It follows from the fact that k is sufficiently large and (25) that
(27)2(p-1p)1/pn∫0h‍du(F(h)-F(u))1/p=0.
Next, we know thatq→-∞ as h→+∞ (F(q)=F(h)). We consider
(28)2(p-1p)1/p(n+1)∫q0‍du(F(q)-F(u))1/p=2(p-1p)1/p(n+1)∫q0‍du(∫uq‍f(t)dt)1/p.
To this end, we have from (H1) that, for any k∈N, there exists Rk∈(0,∞), such that
(29)-f(s)≥kp|s|p-1,s≤-Rk.
Ifq<-Rk, it follows from (28) and (29) that we have that
(30)2(p-1p)1/p(n+1)∫q-Rk‍ds(F(q)-F(s))1/p+2(p-1p)1/p(n+1)∫-Rk0‍ds(F(q)-F(s))1/p=2(p-1p)1/p(n+1)∫q-Rk‍ds(∫sq‍f(w)dw)1/p+2(p-1p)1/p(n+1)∫-Rk0‍ds(∫sq‍f(w)dw)1/p≤2(p-1p)1/p(n+1)∫-Rk0‍(kpp(|q|p-Rkp))-1/pds+2(p-1p)1/p(n+1)∫q-Rk‍(kpp(|q|p-|s|p))-1/pds=2(p-1)1/p(n+1)k{∫-Rk/q1‍1(1-vp)1/pRk[(|q|p-Rkp)]-1/piiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii+∫-Rk/q1‍1(1-vp)1/pdv}=(n+1)k{∫-Rk(p-1)1/p/q(p-1)1/p‍1(1-(vp/p-1))1/p2(p-1)1/pRk[(|q|p-Rkp)]-1/piiiiiiiiiiiiiiiiiiiii+2∫-Rk(p-1)1/p/q(p-1)1/p‍1(1-(vp/(p-1)))1/pdv}⟶(n+1)Ckasq⟶-∞,
where
(31)C≔lim⁡q→-∞{1(1-(vp/(p-1)))1/p2(p-1)1/pRk[(|q|p-Rkp)]-1/piiiiiiiiiiiii+2∫-Rk(p-1)1/p/q(p-1)1/p‍1(1-(vp/(p-1)))1/pdv}=πp.
It follows from the fact that k is sufficiently large and (30) that
(32)2(n+1)∫q0‍du(F(q)-F(u))1/2=0.
Therefore, from (27) and (32), we have that lim⁡h→∞G2n(h)=0.
Proof of (B). Recall that
(33)G2n(h)=2(p-1p)1/p{n∫0h‍du(F(h)-F(u))1/piiiiiiiiiiiiiiiiiiiiiii+(n+1)∫q0‍du(F(q)-F(u))1/p}.
First, let us consider(34)2(p-1p)1/pn∫0h‍du(F(h)-F(u))1/p=2(p-1p)1/pn∫0h‍du(∫uh‍f(t)dt)1/p.
Sincef0∈(0,∞), then, for any ε∈(0,f0/2), there exists δ∈(0,∞) such that
(35)f0-ε≤f(s)sp-1≤f0+ε,0<s<δ.
Thus, if 0<h<δ, the second part of (35) implies that
(36)2n(p-1p)1/p∫0h‍1(∫uh‍f(v)dv)1/pdu≥2n(p-1p)1/p(pf0+ε)1/p∫0h‍1(hp-up)1/pdu=2n(p-1f0+ε)1/p∫0h‍1(hp-up)1/pdu=2n(p-1f0+ε)1/p1(p-1)1/p×∫0(p-1)1/p‍1(1-(up/(p-1)))1/pdu=2n(f0+ε)1/p∫0(p-1)1/p‍1(1-(up/(p-1)))1/pdu=nπp(f0+ε)1/p.
Similarly, from the first part of (35), we have that
(37)2n(p-1p)1/p∫0h‍1(∫uh‍f(v)dv)1/pdu≤2n(p-1p)1/p(pf0-ε)1/p∫0h‍1(hp-up)1/pdu=2n(p-1f0-ε)1/p∫0h‍1(hp-up)1/pdu=2n(p-1f0-ε)1/p1(p-1)1/p×∫0(p-1)1/p‍1(1-(up/(p-1)))1/pdu=2n(f0-ε)1/p∫0(p-1)1/p‍1(1-(up/(p-1)))1/pdu=nπp(f0-ε)1/p.
It follows from (36), (37) and the fact ε is arbitrary that
(38)lim⁡h→02(p-1p)1/pn∫0h‍du(F(h)-F(u))1/p=nπpf01/p.
In fact,q→0 as h→0 (F(h)=F(q)); we consider
(39)2(p-1p)1/p(n+1)∫q0‍du(F(q)-F(u))1/p.
From f0∈(0,∞), then, for any ɛ∈(0,f0/2), there exists δ∈(0,∞) such that
(40)(f0-ɛ)|s|p-1≤-f(s)≤(f0+ɛ)|s|p-1,-δ<s<0.
Thus, if -δ<q<0, the second part of (40) implies that
(41)2(n+1)(p-1p)1/p∫q0‍1(∫uq‍f(v)dv)1/pdu≥2(n+1)(p-1p)1/p(pf0+ε)1/p×∫q0‍1((-q)p-(-u)p)1/pdu=2(n+1)(p-1f0+ε)1/p∫q0‍1((-q)p-(-u)p)1/pdu=2(n+1)(p-1f0+ε)1/p1(p-1)1/p×∫0(p-1)1/p‍1(1-(up/(p-1)))1/pdu=2(n+1)(f0+ε)1/p∫0(p-1)1/p‍1(1-(up/(p-1)))1/pdu=(n+1)πp(f0+ε)1/p.
Similarly, from the first part of (40), we have that
(42)2(n+1)(p-1p)1/p∫q0‍1(∫uq‍f(v)dv)1/pdu≤2(n+1)(p-1p)1/p(pf0-ε)1/p×∫q0‍1((-q)p-(-u)p)1/pdu=2(n+1)(p-1f0-ε)1/p∫q0‍1((-q)p-(-u)p)1/pdu=2(n+1)(p-1f0-ε)1/p1(p-1)1/p×∫0(p-1)1/p‍1(1-(up/(p-1)))1/pdu=2(n+1)(f0-ε)1/p∫0(p-1)1/p‍1(1-(up/(p-1)))1/pdu=(n+1)πp(f0-ε)1/p.
It follows from (41), (42) and the fact that ε is arbitrary that we have that
(43)lim⁡q→02(p-1p)1/p(n+1)∫q0‍du(F(q)-F(u))1/p=(n+1)πpf01/p.
Therefore, from (40) and (43), we have that
(44)lim⁡h→0G2n(h)=(2n+1)πpf01/p.
By analysing G12n(h) defined in (20) instead of G2n(h) in the proof of the above, we have the same result. Thus, we have shown that there are two solutions with 2n interior zeros, which are negative near 0 and positive near 0 for λ∈(0,(2n+1)pπpp/f0), respectively.
Now, in order to achieve the existence ofλ*, we will first establish that G2n′(h)<0 for h large enough. In fact,
(45)G2n′(h)=2(p-1p)1/p×{n∫01‍H(h)-H(hv)(F(h)-F(hv))(p+1)/pdviiiiii-(n+1)dqdh∫01‍H(q)-H(qv)(F(q)-F(qv))(p+1)/pdv}.
First, we consider
(46)n∫01‍H(h)-H(hv)(F(h)-F(hv))(p+1)/pdv,
where H(s)=F(s)-(s/p)f(s),H′(s)=((p-1)/p)f(s)-(s/p)f′(s). From the first part of (H2), it follows that
(47)H(h)-H(hv)≤0,v∈[0,1]
if h is large enough.
Next, let us consider(48)-(n+1)dqdh∫01‍H(q)-H(qv)(F(q)-F(qv))(p+1)/pdv,
where -(dq/dh)>0,H(s)=F(s)-(s/p)f(s),H′(s)=((p-1)/p)f(s)-(s/p)f′(s). From the second part of (H2), we have that H(q)-H(qv)≤0 for v∈[0,1] and |q| large enough. In fact, q→-∞ as h→∞(F(h)=F(q)). Consequently, we get that G2n′(h)<0 for h large enough.
Finally, ifk=2n, this clearly follows by analysing G2n-1(h) defined in (21) instead of G2n(h) in the proof of the case k=2n+1.Proof of Theorem2.
First, we considerk=2n+1.
It follows from the quadrature method that a solution with2n interior zeros exists if for λ>0 there exists h∈(0,∞) such that (λ)1/p=G2n(h). To prove this, we will show that (0,∞)⊂Range(G2n(h)). We achieve this by proving (A1)
lim⁡h→+∞G2n(h)=0,(B1)
lim⁡h→0G2n(h)=∞. The proof of (A1) is the same as the proof of (A) of Theorem 1, so we omit it here; we are only to prove (B1). Recall that
(49)G2n(h)=2(p-1p)1/p{n∫0h‍du(F(h)-F(u))1/piiiiiiiiiiiiiiiiiiiiii+(n+1)∫q0‍du(F(q)-F(u))1/p}.
First let us consider(50)2(p-1p)1/pn∫0h‍du(F(h)-F(u))1/p=2(p-1p)1/pn∫0h‍du(∫uh‍f(t)dt)1/p.
Fromf0=0, then, for any ε>0, there exists δ∈(0,∞) such that
(51)f(s)≤εpsp-1,0<s<δ.
Thus, if 0<h<δ, from (50), we have that
(52)2n(p-1p)1/p∫0h‍1(∫uh‍f(v)dv)1/pdu≥2n(p-1p)1/pp1/pε∫0h‍1(hp-up)1/pdu=2n(p-1)1/pε∫01‍1(1-(u/h)p)1/pd(uh)=2n(p-1)1/pε1(p-1)1/p×∫0(p-1)1/p‍1(1-(sp/(p-1)))1/pds=2nε∫0(p-1)1/p‍1(1-(sp/(p-1)))1/pds=nεπp.
Next, in fact,q→0 as h→0  (F(q)=F(h)); we consider
(53)2(p-1p)1/p(n+1)∫q0‍du(F(q)-F(u))1/p.
Sincef0=0, then, for any ε>0, there exists δ∈(0,∞) such that
(54)-f(s)≤εp|s|p-1,-δ<s<0.
Thus, if -δ<q<0, from (54), we have that
(55)2(p-1p)1/p(n+1)∫q0‍du(∫uq‍f(t)dt)1/p≥2(n+1)(p-1p)1/pp1/pε×∫q0‍1((-q)p-(-u)p)1/pdu=2(n+1)(p-1)1/pε∫01‍1(1-(u/q)p)1/pd(uq)=2(n+1)(p-1)1/pε1(p-1)1/p×∫0(p-1)1/p‍1(1-(vp/(p-1)))1/pdv=2(n+1)ε∫0(p-1)1/p‍1(1-(vp/(p-1)))1/pdv=n+1επp.
From the fact that ε is small and combining (52) and (55), we get that
(56)lim⁡h→0G2n(h)=∞.
By analyzing G12n(h) defined in (20) instead of G2n(h) in the proof of the above, we have the same result. The proof of λ* is similar to the proof of Theorem 1. We omit it here.
Finally, ifk=2n, then it clearly follows by analyzing G2n-1(h) defined in (21) instead of G2n(h) in the proof of the case k=2n+1.Proof of Theorem3.
First, we considerk=2n+1.
It follows from the quadrature method that a solution with2n interior zeros exists if for λ>0 there exists h∈(0,∞) such that (λ)1/p=G2n(h). We prove this by proving (A2)
lim⁡h→∞G2n(h)=0,(B2)
lim⁡h→0G2n(h)=0.The proof of (A2) is the same as the proof of (A) of Theorem 1, so we omit it here; we are only to prove (B2).
Recall that(57)G2n(h)=2(p-1p)1/p{n∫0h‍du(F(h)-F(u))1/piiiiiiiiiiiiiiiiiiiiii+(n+1)∫q0‍du(F(q)-F(u))1/p}.
First, consider(58)2(p-1p)1/pn∫0h‍du(F(h)-F(u))1/p=2(p-1p)1/pn∫0h‍du(∫uh‍f(t)dt)1/p.
Sincef0=∞, then, for any large ϵ, there exists R∈(0,∞) such that
(59)f(s)≥ϵpsp-1,0<s<R.
Thus, if 0<h<R, from (58) and (59), we have that
(60)2(p-1p)1/pn∫0h‍du(∫uh‍f(t)dt)1/p=2n(p-1p)1/p∫0h‍1(∫uh‍f(t)dt)1/pdu≤2n(p-1p)1/pp1/pϵ∫0h‍1(hp-up)1/pdu=2n(p-1)1/pϵ∫01‍1(1-(u/h)p)1/pd(uh)=2n(p-1)1/pϵ1(p-1)1/p×∫0(p-1)1/p‍1(1-(sp/(p-1)))1/pds=2nϵ∫0(p-1)1/p‍1(1-(sp/(p-1)))1/pds=nϵπp.
Next, in fact,q→0 as h→0 (F(q)=F(h)); we consider
(61)2(p-1p)1/p(n+1)∫q0‍du(F(q)-F(u))1/p.
Sincef0=∞, then, for any large ϵ, there exists R∈(0,∞) such that
(62)-f(s)≥ϵp|s|p-1,-R<s<0.
Thus, if-R<q<0, from (62), we have that
(63)2(n+1)(p-1p)1/p∫q0‍1(∫uq‍f(t)dt)1/pdu≤2(n+1)(p-1p)1/pp1/pϵ∫q0‍1((-q)p-(-u)p)1/pdu=2(n+1)(p-1)1/pϵ∫01‍1(1-(u/q)p)1/pd(uq)=2(n+1)ϵ∫0(p-1)1/p‍1(1-(sp/(p-1)))1/pds=n+1ϵπp.
From the fact that ϵ is arbitrary and large and combining (60) and (63), we get that
(64)lim⁡h→0G2n(h)=0.
By analyzing G12n(h) defined in (20) instead of G2n(h) in the proof of the above, we have the same result.
Finally, ifk=2n, then it clearly follows by analyzing G2n-1(h) defined in (20) instead of G2n(h) in the proof of the case k=2n+1.



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*Source: 810193-2014-08-17.xml*