Proof.
We divide the proof into two steps.

Step
1. We define the operator P:C([t0-h,∞),ℝ)→C([t0-h,∞),ℝ) by
(11)(Py)(t)={y(t0)e-β(t-t0) +1Γ(α)∫t0t(t-s)α-1e-β(t-s) ssddss×f(s,ys)ds,t⩾t0,ϕ(t),t∈[t0-h,t0].
The operator P maps BC into itself. Indeed for each y∈BC, and for each t⩾2t0+h, it follows from (H1) that
(12)|(Py)(t)|⩽|y(t0)|e-β(t-t0) +1Γ(α)∫t0t(t-s)α-1e-β(t-s)(f0+l∥ys∥)ds⩽∥ϕ∥e-β(t-t0)+f0+l∥y∥∞Γ(α) ×(∫t0t-(t0+h)(t0+h)α-1e-β(t-s)ds+∫t-(t0+h)t(t-s)α-1ds)⩽∥ϕ∥+(f0+l∥y∥∞)((t0+h)α-1e-β(t0+h)βΓ(α)+(t0+h)αΓ(α+1)).
For each t∈[t0,2t0+h], we have
(13)|(Py)(t)|⩽∥ϕ∥+(f0+l∥y∥∞)(t0+h)αΓ(α+1),
and consequently P(y)∈BC.

Since BC:=BC([t0-h,∞),ℝ) is a Banach space with norm ∥·∥∞, we will show that P:BC→BC is a contraction map. Let y1,y2∈BC. Then, we have for each t⩾t0,
(14)|(Py1)(t)-(Py2)(t)| ⩽1Γ(α)∫t0t(t-s)α-1e-β(t-s) ffffff×|f(s,y1s)-f(s,y2s)|ds ⩽lΓ(α)∫t0t(t-s)α-1e-β(t-s)∥y1s-y2s∥ds.

Therefore, for any t⩾2t0+h,
(15)|(Py1)(t)-(Py2)(t)| ⩽lΓ(α)∥y1(·)-y2(·)∥∞ ×(∫t0t-(t0+h)(t0+h)α-1e-β(t-s)ds dd+∫t-(t0+h)t(t-s)α-1ds) ⩽l((t0+h)α-1e-β(t0+h)βΓ(α)+(t0+h)αΓ(α+1)) ×∥y1(·)-y2(·)∥∞,
and for t0-h⩽t⩽2t0+h,
(16)|(Py1)(t)-(Py2)(t)| ⩽lΓ(α)∥y1(·)-y2(·)∥∞∫t0t(t-s)α-1ds ⩽l(t0+h)αΓ(α+1)∥y1(·)-y2(·)∥∞,
and thus
(17)∥(Py1)(·)-(Py2)(·)∥∞ ⩽l((t0+h)α-1e-β(t0+h)βΓ(α)+(t0+h)αΓ(α+1)) ×∥y1(·)-y2(·)∥∞.
Hence, (10) and (17) imply that the operator P is a contraction. Therefore, P has a unique fixed point by Banach’s contraction principle.

Step
2. For any two solutions x=x(t) and y=y(t) of IVP (1) corresponding to initial values ψ and ϕ, by (4) we can deduce that for all t⩾t0+h and all θ∈[-h,0],
(18)|x(t+θ)-y(t+θ)| ⩽|x(t0)-y(t0)|e-β(t+θ-t0) +1Γ(α)∫t0t+θ(t+θ-s)α-1e-β(t+θ-s) ssssddssss×|f(s,xs)-f(s,ys)|ds ⩽|x(t0)-y(t0)|e-β(t+θ-t0) +lΓ(α)∫t0t+θ(t+θ-s)α-1e-β(t+θ-s)∥xs-ys∥ds.
Then, it follows that
(19)eβt∥xt-yt∥ ⩽|x(t0)-y(t0)|eβ(h+t0) +leβhΓ(α)∫t0t(t-s)α-1eβs∥xs-ys∥ds.
Let w(t)=eβt∥xt-yt∥. Then, we have
(20)w(t)⩽|x(t0)-y(t0)|eβ(h+t0) +leβhΓ(α)∫t0t(t-s)α-1w(s)ds.
Applying Lemma 2, one can see that there exists a constant K such that
(21)w(t) ⩽|x(t0)-y(t0)|eβ(h+t0) +KleβhΓ(α)∫t0t(t-s)α-1|x(t0)-y(t0)|eβ(h+t0)ds ⩽|x(t0)-y(t0)|eβ(h+t0)(1+KleβhΓ(α+1)(t-t0)α).
Hence, we obtain
(22)eβt∥xt-yt∥=w(t)⩽|x(t0)-y(t0)|eβ(h+t0) ×(1+KleβhΓ(α+1)(t-t0)α),
and thus for all t⩾t0+h,
(23)|x(t)-y(t)| ⩽|x(t0)-y(t0)|e-β(t-h-t0)(1+KleβhΓ(α+1)(t-t0)α),
which implies that the solutions of IVP (1) are uniformly asymptotically stable.

Now we give global existence and uniform asymptotic stability results based on the nonlinear alternative of Leray-Schauder type.

Proof.
Let P:C([t0-h,∞),ℝ)→C([t0-h,∞),ℝ) be defined as in (11). First, we show that P maps BC into itself. Let K1=supt⩾t0k1(t), K2=supt⩾t0k2(t). Indeed, the map P(y) is continuous on [t0-h,+∞) for each y∈BC, and for each t⩾2t0+h, (H2) implies that
(26)|(Py)(t)| ⩽|y(t0)|e-β(t-t0) +1Γ(α)∫t0t(t-s)α-1e-β(t-s)(K1+K2∥ys∥)ds ⩽∥ϕ∥e-β(t-t0)+K1+K2∥y∥∞Γ(α) ×(∫t0t-(t0+h)(t0+h)α-1e-β(t-s)ds ssss+∫t-(t0+h)t(t-s)α-1ds) ⩽∥ϕ∥+(K1+K2∥y∥∞) ×((t0+h)α-1e-β(t0+h)βΓ(α)+(t0+h)αΓ(α+1)),
for each t∈[t0,2t0+h], we have
(27)|(Py)(t)|⩽∥ϕ∥+(K1+K2∥y∥∞)(t0+h)αΓ(α+1),
and for any t∈[t0-h,t0],
(28)|(Py)(t)|⩽∥ϕ∥.
Thus,
(29)∥P(y)∥∞ ⩽∥ϕ∥+(K1+K2∥y∥∞) ×((t0+h)α-1e-β(t0+h)βΓ(α)+(t0+h)αΓ(α+1)),
and consequently P(y)∈BC.

Next, we show that the operator P is continuous and completely continuous, and there exists an open set U⊂BC with y≠λP(y) for λ∈(0,1) and y∈∂U.

Step
1 (
P
is continuous). Let {yn} be a sequence such that yn→y in BC. Then, there exist R>0 and N>0 such that
(30)∥yn∥∞+∥y∥∞<R, ∀n⩾N.
Let ε>0 be given. Since (H4) holds, there is a real number T>0 such that
(31)2Γ(α)∫t0t(t-s)α-1e-β(t-s)(k1(s)+k2(s)R)ds<ε,
for all t⩾T. Now we consider the following two cases.

Case
1. If t⩾T, then it follows from (H3) and (30)-(31) that for n sufficiently large
(32)|Pyn(t)-Py(t)| ⩽|yn(t0)-y(t0)|e-β(t-t0) +1Γ(α)∫t0t(t-s)α-1e-β(t-s)|f(s,yns)-f(s,ys)|ds ⩽|yn(t0)-y(t0)| +2Γ(α)∫t0t(t-s)α-1e-β(t-s) dddddddd×(k1(s)+k2(s)R)ds<2ε.

Case 2. If t0⩽t⩽T, since f is a continuous function, one has
(33)|Pyn(t)-Py(t)| ⩽|yn(t0)-y(t0)| +1Γ(α)∫t0t(t-s)α-1e-β(t-s)|f(s,yns)-f(s,ys)|ds ⩽|yn(t0)-y(t0)| +(T-t0)αΓ(α+1)sups∈[t0,T]|f(s,yns)-f(s,ys)|.
Note that yn→y in BC. Hence, (32) and (33) imply that
(34)∥P(yn)-P(y)∥∞⟶0 as n⟶∞.

Step
2 (P maps bounded sets into bounded sets in
B
C
). Indeed, it is enough to show that for any η>0, there exists a positive constant ℓ such that for each y∈Bη={y∈BC:∥y∥∞⩽η} one has ∥P(y)∥∞⩽ℓ. Let y∈Bη. Then, we have for each t⩾2t0+h,
(35)|(Py)(t)|⩽|y(t0)|e-β(t-t0) +1Γ(α)∫t0t(t-s)α-1e-β(t-s)|f(s,ys)|ds⩽η+K1+K2∥y∥∞Γ(α)∫t0t(t-s)α-1e-β(t-s)ds⩽η+(K1+K2η) ×((t0+h)α-1e-β(t0+h)βΓ(α)+(t0+h)αΓ(α+1))=:ℓ,
and for each t with t0⩽t⩽2t0+h,
(36)|(Py)(t)|⩽η+(K1+K2η)(t0+h)αΓ(α+1).

Hence, ∥P(y)∥∞⩽ℓ.

Step
3 (P maps bounded sets into equicontinuous sets on every compact subset [t0-h,b] of [t0-h,∞)). Let t1,t2∈[t0,b], t1<t2, and let Bη be a bounded set of BC as in Step 2. Let y∈Bη. Then, we have
(37)|(Py)(t2)-(Py)(t1)| ⩽|y(t0)e-β(t2-t0)-y(t0)e-β(t1-t0)| +1Γ(α)∫t0t1|((t2-s)α-1e-β(t2-s) ddddddddddd-(t1-s)α-1e-β(t1-s))f(s,ys)|ds +1Γ(α)∫t1t2|(t2-s)α-1e-β(t2-s)f(s,ys)|ds ⩽|y(t0)|eβt0|e-βt2-e-βt1|+K1+K2ηΓ(α+1)(t2-t1)α +K1+K2ηΓ(α)∫t0t1((t1-s)α-1e-β(t1-s) ddddddddddddd-(t2-s)α-1e-β(t2-s))ds ⩽|y(t0)|eβt0|e-βt2-e-βt1|+K1+K2ηΓ(α+1)(t2-t1)α +K1+K2ηΓ(α)∫t0t1((t1-s)α-1e-β(t1-s) ddddddddddddd-(t2-s)α-1e-β(t1-s))ds +K1+K2ηΓ(α)∫t0t1((t2-s)α-1e-β(t1-s) ssssssssssshsssdh-(t2-s)α-1e-β(t2-s))ds.

Observing that
(38)K1+K2ηΓ(α) ×∫t0t1((t1-s)α-1e-β(t1-s)-(t2-s)α-1e-β(t1-s))ds ⩽K1+K2ηΓ(α)∫t0t1((t1-s)α-1-(t2-s)α-1)ds ⩽K1+K2ηΓ(α+1)((t1-t0)α-(t2-t0)α+(t2-t1)α) ⩽K1+K2ηΓ(α+1)(t2-t1)α,
from Taylor’s theorem, we obtain
(39)K1+K2ηΓ(α) ×∫t0t1((t2-s)α-1e-β(t1-s)-(t2-s)α-1e-β(t2-s))ds ⩽K1+K2ηΓ(α)(t2-t1)α-1∫t0t1(e-β(t1-s)-e-β(t2-s))ds ⩽K1+K2ηβΓ(α)(t2-t1)α-1(1-e-β(t2-t1)) =K1+K2ηΓ(α)((t2-t1)α+o(t2-t1)t2-t1(t2-t1)α),
where limt2-t1→0(o(t2-t1)/(t2-t1))=0. By (37)–(39), we can conclude that
(40)|(Py)(t2)-(Py)(t1)| ⩽ηeβt0|e-βt2-e-βt1| +2(K1+K2η)Γ(α+1)(t2-t1)α +K1+K2ηΓ(α)((t2-t1)α+o(t2-t1)t2-t1(t2-t1)α).
As t1→t2, the right-hand side of the above inequality tends to zero. The equicontinuity for the cases t1<t2⩽t0 and t1⩽t0⩽t2 is obvious.

Step
4 (P maps bounded sets into equiconvergent sets). Let y∈Bη. Then
(41)|(Py)(t)| ⩽|y(t0)|e-β(t-t0) +1Γ(α)∫t0t(t-s)α-1e-β(t-s)|f(s,ys)|ds ⩽ηe-β(t-t0) +1Γ(α)∫t0t(t-s)α-1e-β(t-s)(k1(s)+k2(s)η)ds.
Therefore, (H4) implies that |(Py)(t)| uniformly (with respect to y∈B(η)) converges to 0 as t→∞. As a consequence of Steps 1–4, we can conclude that P:BC→BC is continuous and completely continuous.

Step
5 (a priori bounds). We now show that there exists an open set U⊆BC with y≠λP(y) for λ∈(0,1) and y∈∂U.

Let y∈BC and y=λP(y) for some 0<λ<1. Then, for each t∈[t0,∞), we obtain
(42)y(t)=λ[1Γ(α)∫t0ty(t0)e-β(t-t0) dddd+1Γ(α)∫t0t(t-s)α-1e-β(t-s)f(s,ys)ds].
By (H3), we have that for all θ∈[-h,0] and t⩾t0+h,
(43)|y(t+θ)| ⩽|y(t0)|e-β(t+θ-t0) +1Γ(α)∫t0t+θ(t+θ-s)α-1e-β(t+θ-s)|f(s,ys)|ds ⩽|y(t0)|e-β(t+θ-t0) +1Γ(α)∫t0t+θ(t+θ-s)α-1e-β(t+θ-s)(K1+K2∥ys∥)ds,
and thus
(44)∥yt∥ ⩽|y(t0)|e-β(t-h-t0) +1Γ(α)∫t0t(t-s)α-1e-β(t-h-s)(K1+K2∥ys∥)ds.
From the arguments in (26)-(27), we can conclude that for each t∈[t0,∞),
(45)1Γ(α)∫t0t(t-s)α-1e-β(t-s)ds ⩽(t0+h)α-1e-β(t0+h)βΓ(α)+(t0+h)αΓ(α+1)=:R1.
Hence,
(46)eβt∥yt∥ ⩽∥ϕ∥eβ(h+t0)+eβ(h+t)K1R1 +K2eβhΓ(α)∫t0t(t-s)α-1eβs∥ys∥ds.
Let R2=∥ϕ∥eβ(h+t0)+eβhK1R1. Then, from Lemma 2, there exists K such that we have for all t⩾t0+h,
(47)∥yt∥⩽R2+KK2R2eβhΓ(α+1)(t-t0)αe-βt.
Since limt→∞(t-t0)αe-βt=0, there exists R3>0 such that
(48)∥y∥∞⩽R3.

Set
(49)U={y∈BC:∥y∥∞<R3+1}.P:U→BC is continuous and completely continuous. From the choice of U, there is no y∈∂U such that y=λP(y), for λ∈(0,1). As a consequence of Leray-Schauder fixed-point theorem, we deduce that P has a fixed point y in U.

Step
6 (uniform asymptotic stability of solutions). Let B⊂C([-h,0],ℝ) be bounded; that is, there exists d⩾0 such that
(50)∥ψ∥=supθ∈[-h,0]|ψ(θ)|⩽d ∀ψ∈B.
From the similar arguments in Step 4, we can deduce that there exists R4>0 such that for all solutions y(t,t0,ϕ) of IVP (1) with initial data ϕ∈B, we have
(51)∥y∥∞⩽R4, ∀ϕ∈B.

Now we consider two solutions x=x(t) and y=y(t) of IVP (1) corresponding to the initial values ψ and ϕ. Note that for all t⩾t0,
(52)|x(t)-y(t)| ⩽|x(t0)-y(t0)|e-β(t-t0) +1Γ(α)∫t0t(t-s)α-1e-β(t-s)(|f(s,xs)|+|f(s,ys)|)ds ⩽2de-β(t-t0) +2Γ(α)∫t0t(t-s)α-1e-β(t-s)(k1(s)+k2(s)R4)ds.
Then, the proof of uniform asymptotic stability of solutions can be done by making use of (H4) and (52).

The proof of Theorem 5 is completed.