We present the following two assumptions.

Proof.
The proof is divided into four steps.

Step 1. We show that Tλ is well defined on P and Tλ(P)⊂P, and Tλ is decreasing in x.

In fact, for any x∈P, by the definition of P, there exists two positive numbers 0<lx<1, Lx>1 such that lxe(t)≤x(t)≤Lxe(t) for any t∈[0,1]. It follows from Lemma 8 and (H1)-(H2) that
(32)(Tλx)(t)=λb∫01G(t,s)(∫0s(s-τ)β-1f(τ,x(τ))dτ)q-1ds≤λbe(t)∫01σ2(s)×(∫0s(s-τ)β-1f(τ,lxe(τ))dτ)q-1ds<+∞.

Now take c=maxt∈[0,1]x(t), by (H2), for any s∈(0,1), f(s,c)≢0. Thus by the continuity of f(t,x) and Lemma 8 and (32), we have
(33)(Tλx)(t)≥λbe(t)∫01σ1(s)×(∫0s(s-τ)β-1f(τ,x(τ))dτ)q-1ds≥λbe(t)∫01σ1(s)(∫0s(s-τ)β-1f(τ,c)dτ)q-1ds>0, t∈(0,1).

Take
(34)lx′=min{1,λb∫01σ1(s)(∫0s(s-τ)β-1f(τ,c)dτ)q-1ds},Lx′=max{(∫0s(s-τ)β-1f(τ,lxe(τ))dτ)1,λb ×∫01σ2(s)(∫0s(s-τ)β-1f(τ,lxe(τ))dτ)q-1ds},
then by (32) and (33),
(35)lx′e(t)≤(Tλx)(t)≤Lx′e(t),
which implies that Tλ is well defined and Tλ(P)⊂P. And the operator Tλ is decreasing in x from (H1). Moreover, by direct computations, we also have
(36)-𝒟tβ(φp(𝒟tα(Tλx)))(t)=λf(t,x(t)), t∈(0,1),(Tλx)(0)=0, 𝒟tα(Tλx)(0)=0, 𝒟tγ(Tλx)(1)=∑j=1m-2aj𝒟tγ(Tλx)(ξj).

Step 2. In this step, we will focus on lower and upper solutions of the fractional boundary value problem (4).

By Lemma 8, we have
(37)b∫01G(t,s)(∫0s(s-τ)β-1f(τ,e(τ))dτ)q-1dsa ≥e(t)b∫01σ1(s)(∫0s(s-τ)β-1f(τ,e(τ))dτ)q-1ds,e(t)b∫01(∫0s(s-τ)β-1f(τ,e(τ))dτ)q-1∀t∈[0,1].
Let
(38)μ=(b∫01σ1(s)(∫0s(s-τ)β-1f(τ,e(τ))dτ)q-1ds)-1;
it follows from (37) that
(39)μb∫01G(t,s)(∫0s(s-τ)β-1f(τ,e(τ))dτ)q-1ds≥e(t), (∫0s(s-τ)β-1(τ,e(τ))dτ)q-1∀t∈[0,1].

On the other hand, take
(40)ν(t)=b∫01G(t,s)(∫0s(s-τ)β-1f(τ,e(τ))dτ)q-1ds;
then by monotonicity of f in x and (37)–(40), for any λ>μ, we have
(41)∫01G(t,s)(∫0s(s-τ)β-1f(τ,λν(τ))dτ)q-1ds ≤∫01G(t,s)(∫0s(s-τ)β-1f(τ,μν(τ))dτ)q-1ds ≤∫01σ2(s)(∫0s(s-τ)β-1f(τ,e(τ))dτ)q-1ds <+∞.
From (S1), we have
(42)limκ→+∞κ1/(q-1)f(t,κx)=+∞,
uniformly on (t,x)∈(0,1)×(0,∞). Thus there exists large enough λ*>μ>0, such that, for any t∈(0,1),
(43)λ*1/(q-1)f(s,λ*e(s))≥βb(∫01σ1(s)sβ(q-1)ds)-1/(q-1),
which yields
(44)λ*b∫01G(t,s)(∫0s(s-τ)β-1f(τ,λ*e(τ))dτ)q-1ds≥β(∫01σ1(s)sβ(q-1)ds)-1/(q-1)a×∫01G(t,s)(∫0s(s-τ)β-1dτ)q-1ds≥β(∫01σ1(s)sβ(q-1)ds)-1/(q-1)ai×∫01σ1(s)(∫0s(s-τ)β-1dτ)q-1dse(t)=e(t),×∫01σ1(s)(∫0s(s-τ)β-1dτ)q-1ghjhg∀t∈[0,1].
Letting
(45)ϕ(t)=λ*b∫01G(t,s)(∫0s(s-τ)β-1f(τ,e(τ))dτ)q-1ds=λ*ν(t)=Tλ*e,ψ(t)=λ*b∫01G(t,s)(∫0s(s-τ)β-1f(τ,λ*ν(τ))dτ)q-1ds =Tλ*ϕ,
and by Lemma 9, (39), (44), and (45), one has
(46)ϕ(t)=λ*b∫01G(t,s)(∫0s(s-τ)β-1f(τ,e(τ))dτ)q-1ds≥e(t),ϕ(0)=0, 𝒟tϕϕ(0)=0, 𝒟tγ(ϕx)(1)=∑j=1m-2aj𝒟tγϕ(ξj),(47)ψ(t)=λ*b∫01G(t,s)(∫0s(s-τ)β-1f(τ,λ*ν(τ))dτ)q-1ds≥e(t),ψ(0)=0, 𝒟tϕψ(0)=0, 𝒟tγ(ψx)(1)=∑j=1m-2aj𝒟tγψ(ξj).

By Step 1 and (46), (47), we know ϕ(t),ψ(t)∈P. And it follows from (45)–(47) that
(48)e(t)≤ψ(t)≤ϕ(t), ∀t∈[0,1].
Consequently, it follows from (44)–(48) that
(49)𝒟tβ(φp(𝒟tαψ))(t)+λ*f(t,ψ(t))=𝒟tβ(φp(𝒟tα(Tλ*ϕ)))(t)+λ*f(t,ψ(t))=-λ*f(t,ϕ(t))+λ*f(t,ψ(t))≥0,𝒟tβ(φp(𝒟tαϕ))(t)+λ*f(t,ϕ(t))=𝒟tβ(φp(𝒟tα(Tλ*e)))(t)+λ*f(t,ϕ(t))=-λ*f(t,e(t))+λ*f(t,ϕ(t))≤0;
that is, ϕ(t) and ψ(t) are a couple of lower and upper solutions of fractional boundary value problem (4) by (46)–(49), respectively.

Step 3. Let
(50)F(t,x)={f(t,ψ(t)),x<ψ(t),f(t,x(t)),ψ(t)≤x≤ϕ(t),f(t,ϕ(t)), x>ϕ(t).
It follows from (H1) and (46) that F:(0,1)×[0,+∞)→[0,+∞) is continuous.

We will show that the fractional boundary value problem
(51)-𝒟tβ(φp(𝒟tαx))(t)=λ*F(t,x(t)), t∈(0,1),x(0)=0, 𝒟tαx(0)=0, 𝒟tγx(1)=∑j=1m-2aj𝒟tγx(ξj)
has a positive solution.

To see this, we consider the operator Aλ*:C[0,1]→C[0,1] defined as follows:
(52)(Aλ*x)(t)=λ*b∫01G(t,s)×(∫0s(s-τ)β-1F(τ,x(τ))dτ)q-1ds,∫0s(s-τ)β-1τ,x(τ)sdggdsit∈[0,1].
Obviously, a fixed point of the operator Aλ* is a solution of the BVP (51). Noting that ϕ∈P, then there exists a constant 0<lϕ<1 such that ϕ(t)≥lϕe(t), t∈[0,1]. Thus for all x∈E, it follows from Lemma 8, (50), and (H2) that
(53)(Aλ*x)(t)= ≤λ*b∫01σ2(s)(∫0s(s-τ)β-1F(τ,x(τ))dτ)q-1ds= ≤λ*b∫01σ2(s)(∫0s(s-τ)β-1f(τ,ϕ(τ))dτ)q-1ds= ≤λ*b∫01σ2(s)(∫0s(s-τ)β-1f(τ,lϕe(τ))dτ)q-1ds= <+∞,
which implies that the operator Aλ* is uniformly bounded.

From the uniform continuity of G(t,s) and the Lebesgue dominated convergence theorem, we easily obtain that A is equicontinuous. Thus by the means of the Arzela-Ascoli theorem, we have that Aλ*:E→E is completely continuous. The Schauder fixed point theorem implies that Aλ* has at least a fixed point w such that w=Aλ*w.

Step 4. We will prove that the boundary value problem (4) has at least one positive solution.

In fact, we only need to prove that
(54)ψ(t)≤w(t)≤ϕ(t), t∈[0,1].
By (46), (47) and noticing that w is fixed point of Aλ*, we know that
(55)ϕ(0)=0, 𝒟tγϕ(1)=∑j=1m-2aj𝒟tγϕ(ξj), 𝒟tαϕ(0) =0,w(0)=0, 𝒟tγw(1)=∑j=1m-2aj𝒟tγw(ξj), 𝒟tαw(0)=0.
Notice that the definition of F and the function f(t,x) is nonincreasing in x, we obtain
(56)f(t,ϕ(t))≤F(t,x(t))≤f(t,ψ(t)), ∀x∈E.
So by (48) and (56),
(57)f(t,ϕ(t))≤F(t,x(t))≤f(t,e(t)), ∀x∈E.
Thus one has by (57)
(58)𝒟tβ(φp(𝒟tαϕ))(t)-𝒟tβ(φp(𝒟tαw))(t)=𝒟tβ(φp(𝒟tαϕ)-φp(𝒟tαw))(t)=-λ*f(t,e(t))+λ*F(t,w(t))≤0,=-λ*f(t,e(t))+λ*dfgd∀t∈[0,1].
Let z(t)=φp(𝒟tαϕ(t))-φp(𝒟tαw(t)); then
(59)𝒟tβz(t)≤0, t∈[0,1],
and (55) implies that z(0)=0. It follows from (21) that
(60)z(t)≤0,
and then
(61)φp(𝒟tαϕ(t))-φp(𝒟tαw(t))≤0.
Notice that φp is monotone increasing; we have
(62)𝒟tαϕ(t)≤𝒟tαw(t), that is, 𝒟tα(ϕ-w)(t)≤0.
It follows from Lemma 10 and (55) that
(63)ϕ(t)-w(t)≥0.
Thus we have w(t)≤ϕ(t) on [0,1]. By the same way, we also have w(t)≥ψ(t) on [0,1]. So
(64)ψ(t)≤w(t)≤ϕ(t), t∈[0,1].
Consequently, F(t,w(t))=f(t,w(t)), t∈[0,1]. Then w(t) is a positive solution of the problem (4).

Finally, by (48) and (64) and ϕ∈P, we have
(65)e(t )≤ψ(t)≤w(t)≤ϕ(t)≤lϕe(t)=ne(t),
where
(66)n=lϕ>1.