Proof.
For convenience, we still denote φ(0)=x. Select δ>0 and construct set
(3.1)V={(u,v)∣‖u-x‖α≤δ, ‖v-x‖α≤δ}.
Let B=∥f(x,x)∥, choose sufficient small T<τ such that
(3.2)‖(e-At-I)x‖α≤δ2, 0≤t<T,(3.3)Cα(B+2Lδ)∫0Tu-αe-audu≤δ2.
Let Y be the Banach space C([-τ,T];X) with the usual supremum norm which we denote by ∥·∥Y. Let S be the nonempty closed and bounded subset of Y defined by
(3.4)S={y:y∈Y, ‖y(t)-Aαx‖≤δ}.
On S we define a mapping F by
(3.5)Fy(t)={e-tAAαx+∫0tAαe-(t-s)Af(A-αy(s),A-αy(s-τ))ds,0<t<T,Aαφ(t),-τ≤t≤0.
Next we will utilize the contraction mapping theorem to prove the existence of fixed point. In order to complete this work, we need to verify that F maps S into itself and F is a contraction mapping on S with the contraction constant ≤1/2.

It is easy to see from (3.4) and (3.5) that for -τ≤t≤0, F:S→S. For 0<t<T, considering (2.1), (2.3), (3.2), and (3.3), we obtain
(3.6)‖Fy(t)-Aαx‖≤‖e-tAAαx-Aαx‖+∫0t‖Aαe-(t-s)Af(A-αy(s),A-αy(s-τ))‖ds≤‖e-tAAαx-Aαx‖+∫0t‖Aαe-(t-s)Af(A-αy(s),A-αy(s-τ))-f(x,x)‖ds+∫0t‖Aαe-(t-s)Af(x,x)‖ds≤δ2+Cα(2Lδ+B)∫0t(t-s)-αe-a(t-s)ds≤δ2+Cα(2Lδ+B)∫0Tu-αe-audu≤δ.
Therefore F:S→S. Furthermore if y1,y2∈S then from (3.3) and (3.5)
(3.7)‖Fy1(t)-Fy2(t)‖ ≤∫0t‖Aαe-(t-s)A[f(A-αy1(s),A-αy1(s-τ))-f(A-αy2(s),A-αy2(s-τ))]‖ds ≤∫0t‖Aαe-(t-s)A‖‖[f(A-αy1(s),A-αy1(s-τ))-f(A-αy2(s),A-αy2(s-τ))]‖ds ≤∫0tCα(t-s)-αe-a(t-s)L‖y1(s)-y2(s)‖ds ≤CαL∫0Tu-αe-auds⋅‖(y1-y2)‖Y ≤12‖(y1-y2)‖Y,
which implies that
(3.8)‖Fy1(t)-Fy2(t)‖Y≤12‖(y1-y2)‖Y.
By the contraction mapping theorem the mapping F has a unique fixed point y∈S. This fixed point satisfies the following:
(3.9)y(t)=e-tAAαx+∫0tAαe-(t-s)Af(A-αy(s),A-αy(s-τ))ds, 0<t<T,(3.10)y(t)=Aαφ(t), -τ≤t≤0.

From (2.1) and the continuity of y it follows that t→f(A-αy(t),A-αy(t-τ)) is continuous on [0,T] and a fortiori bounded on this interval. Let
(3.11)‖f(A-αy(t),A-αy(t-τ))‖≤N.
Next we want to show that t→f(A-αy(t),A-αy(t-τ)) is locally Hölder continuous on (0,T). To this end, we show first that the solution y of (3.9) is locally Hölder continuous on (0,T).

Select [t0,t1]⊂(0,T), t0≤t<t+h≤t1 such that
(3.12)‖y(t+h)-y(t)‖≤‖(e-hA-I)Aαe-tAx‖+∫0t‖(e-hA-I)Aαe-(t-s)Af(A-αy(s),A-αy(s-τ))‖ds+∫tt+h‖Aαe-(t+h-s)Af(A-αy(s),A-αy(s-τ))‖ds=I1+I2+I3.
Considering (2.3) and (2.4), we select β∈(0,1-α) such that
(3.13)I1≤Cβhβ‖Aα+βe-tAx‖≤CβhβCα+βt-(α+β)‖x‖≤M1hβ,I2≤NCβhβ∫0t‖Aα+βe-(t-s)A‖ds≤NChβ∫0t(t-s)-(α+β)ds≤M2hβ,I3≤NCα∫tt+h(t+h-s)-αds=NCα1-αh1-α≤M3hβ.
Synthesizing (3.12) and (3.13), we get
(3.14)‖y(t+h)-y(t)‖≤Chβ, t∈[t0,t1]⊂(0,T).

So we proved the solution y of (3.9) is locally Hölder continuous on (0,T). Furthermore, in view of (2.1) we have
(3.15)‖f[A-αy(t+h),A-αy(t+h-τ)]-f[A-αy(t),A-αy(t-τ)]‖ ≤L(‖y(t+h)-y(t)‖+‖y(t+h-τ)-y(t-τ)‖) ≤LChβ+‖Aαφ(t+h-τ)-Aαφ(t-τ)‖ ≤Mhγ.

Let y be the solution of (3.9) and (3.10) and f~(t)=f(A-αy(t),A-αy(t-τ)). In view of locally Hölder continuous on (0,T) of f~(t), consider the inhomogeneous initial value problem
(3.16)u′(t)+Au=f~(t), 0<t<T,u(0)=x.
By Corollary 4.3.3 in [1], this problem has a unique solution and the solution is given by
(3.17)u(t)=e-tAx+∫0te-(t-s)Af(A-αy(s),A-αy(s-τ))ds.
Each term of (3.17) is in D(A) and a fortiori in D(Aα). Operating on both sides of (3.17) with Aα we find
(3.18)Aαu(t)=e-tAAαx+∫0tAαe-(t-s)Af(A-αy(s),A-αy(s-τ))ds.
By (3.9) the right-hand side of (3.18) equals y(t) and therefore u(t)=A-αy(t). So for 0<t<T, by (3.17) we have
(3.19)u(t)=e-tAx+∫0te-(t-s)Af(u(s),u(s-τ))ds.
So u is a u∈C1(0,T;X) solution of (1.4). The uniqueness of u follows readily from the uniqueness of the solutions of (3.9) and (3.16), and the proof is complete.

Before giving our global existence theorem, we should first prove extended theorem of solution.

Proof.
Suppose Tmax <+∞, there exists a closed bounded B subset of U and τ0<Tmax such that for τ0≤t<Tmax u(t)∈B. We prove there exists x*∈B such that
(3.20)limt→Tmax-u(t)=x*
in Xα, which implies the solution may be extended beyond time Tmax .

Now let
(3.21)C=sup{‖f(u,v)‖,(u,v)∈B}.
We show firstly that ∥u(t)∥β remains bounded as t→Tmax - for any β∈[α,1).

Observe that if α≤β<1, τ0≤t<Tmax , in view of (2.3) and (3.19) we have
(3.22)‖u(t)‖β≤‖Aβ-αe-tA‖‖x‖α+∫0t‖Aβe-(t-s)A‖‖f(u(s),u(s-τ))‖ds≤Cβ-αt-(β-α)‖x‖α+CβC∫0t(t-s)-βds=Cβ-αt-(β-α)‖x‖α+CβC1-βt1-β≤M, 0<τ0≤t<Tmax.

Secondly, suppose τ0≤t1<t<Tmax , so
(3.23)u(t)-u(t1)=(e-(t-t1)A-I)u(t1)+∫t1te-(t-s)Af(u(s),u(s-τ))ds.
From (2.3) and (2.4) we get
(3.24)‖u(t)-u(t1)‖α≤‖(e-(t-t1)A-I)Aαu(t1)‖+C∫t1t‖Aαe-(t-s)A‖ds≤Cβ-α(t-t1)β-α‖Aβ-α+αu(t1)‖+CCα∫t1t(t-s)-αds=Cβ-α(t-t1)β-α‖u(t1)‖β+CCα1-α(t-t1)1-αds≤C0(t-t1)β-α.
Thus (3.20) holds, and the proof is completed.

Proof.
We need to verify that ∥u(t)∥α is bounded when t→Tmax -. As for 0≤t<Tmax (3.26)u(t)=e-tAx+∫0te-(t-s)Af(u(s),u(s-τ))ds.
Considering (3.25), we can obtain
(3.27)‖u(t)‖α=‖Aαu(t)‖≤‖e-tAAαx‖+∫0t‖Aαe-(t-s)A‖L(‖u(s)‖α+‖u(s-τ)‖α)ds≤C1‖x‖α+LCα∫0t(t-s)-α‖u(s)‖αds+LCα∫0t(t-s)-α‖u(s-τ)‖αds,
For
(3.28)∫0t(t-s)-αu(s-τ)ds=s-τ=w∫-τt-τ(t-τ-w)-αu(w)dw.

Case 1. If Tmax ≤τ. Because u(t)=φ(t) for -τ≤t≤0 and φ is Hölder continuous from [-τ,0] to Xα. Let
(3.29)M=maxt∈[-τ,0]‖φ(t)‖,(3.30)∫0t(t-s)-α‖u(s-τ)‖αds=∫-τt-τ(t-τ-w)-α‖φ(w)‖αdw≤M∫-τ0(t-τ-w)-αdw≤M1-αt1-α≤M1, t∈[0,Tmax).
From (3.27), we immediately get
(3.31)‖u(t)‖α≤a+b∫0t(t-s)-α‖u(s)‖αds.
From Lemma 2.3, that is, Gronwall’s inequality, we find ∥u(t)∥α≤C.

Case 2. If Tmax >τ, still let
(3.32)M=maxt∈[-τ,0]‖φ(t)‖,
because
(3.33)∫0t(t-s)-α‖u(s-τ)‖αds=∫-τt-τ(t-τ-w)-α‖u(w)‖αdw=∫-τ0(t-τ-w)-α‖φ(w)‖αdw+∫0t-τ(t-τ-w)-α‖u(w)‖αdw≤M∫-τ0(t-τ-w)-αdw+∫0t(t-s)-α‖u(s)‖αds≤M0+∫0t(t-s)-α‖u(s)‖αds.
From (3.27) again, we obtain
(3.34)‖u(t)‖α≤a0+b0∫0t(t-s)-α‖u(s)‖αds.
By Gronwall’s inequality again, we get ∥u(t)∥α≤C. This completes the proof of this theorem.