1 Introduction

Recently, fractional differential systems have played an important role in physics, chemistry, engineering, biology, finance etc., due to the memory character of fractional derivative, which is a generalization of integer-order derivative and can describe many phenomena that an integer derivative cannot characterize (see [1–7] and the references therein).

In this paper, we study the existence of S-asymptotically T-periodic solutions for the following fractional differential equation on a Banach space X:

$$\begin{aligned} \textstyle\begin{cases} {}^{c} D_{t}^{q} u(t)= Au(t) +f(t, u_{t}),\quad t> 0,\\ u(t)=\phi(t), \quad t\in[-\delta, 0], \end{cases}\displaystyle \end{aligned}$$
(1.1)

where \(q\in(0, 1)\) and \(\delta>0\). The fractional derivative is understood here in the Caputo sense. A is an almost sectorial operator to be introduced later. \(u_{t}: [-\delta, 0]\rightarrow X\) is defined by \(u_{t}(\theta)=u(t+\theta)\) for \(\theta\in[-\delta, 0]\). f is a function to be specified later. \(\phi\in C([-\delta, 0] , X)\).

The literature concerning S-asymptotically T-periodic functions with values in Banach spaces is very new. There are some papers dealing with the existence of S-asymptotically T-periodic solutions of differential equations and fractional differential equations with a sectorial operator in finite as well as infinite dimensional spaces (cf. [4–6, 8–14]). However, von Wahl first introduced examples of almost sectorial operators which are not sectorial [15]. To the best of the authors’ knowledge, there are few papers on the existence of S-asymptotically T-periodic (mild) solutions for fractional differential equation with almost sectorial operator of type (1.1).

We will now present a summary of this work. In Section 2, we recall some fundamental properties of S-asymptotically T-periodic functions and preliminary facts. The existence and uniqueness of S-asymptotically T-periodic mild solution of problem (1.1) are discussed in Section 3, and an example is given to illustrate our result.

2 Preliminaries

Let \((X, \Vert \cdot \Vert )\) be a Banach space and \(L(X)\) be the space of all bounded linear operators from X to X with the usual operator norm \(\Vert x\Vert _{L(X)}\). \(C_{b}(\mathbb{R}_{+}, X)\) denotes the space of the continuous bounded functions from \([0, +\infty)\) to X, endowed with the norm \(\Vert x\Vert _{\infty}=\sup_{t\geq0}\Vert x(t)\Vert \). \(C([-\delta, 0], X)\) denotes the space of the continuous functions from \([-\delta, 0]\) to X with the norm \(\Vert x\Vert _{[-\delta, 0]} =\sup_{t \in [-\delta, 0]} \Vert x(t)\Vert \). \(L^{p}((0, +\infty), (0, +\infty))\) is the \(L^{p}\) space with the norm \({\Vert x\Vert _{p}= (\int_{0}^{+\infty} \vert x(t)\vert ^{p} \,dt )^{\frac{1}{p}}}\) for \(1\leq p<\infty\), and we abbreviate this notation to \(L^{p}\).

We recall the following basic definitions of the fractional calculus theory. For more details, we refer to [1, 2].

Definition 2.1

[1, 2]

The fractional integral of order q with the lower limit zero for a function \(g \in L^{1}[0, \infty)\) is defined as

$$I_{t}^{q}g(t)=\frac{1}{\Gamma(q)} \int_{0}^{t} (t-s)^{q-1}g(s)\,ds,\quad t>0, 0< q< 1, $$

where \(\Gamma(\cdot)\) is the gamma function.

Definition 2.2

[1, 2]

The Caputo derivative of order q for a function \(g \in C^{1}[0, \infty)\) can be written as

$${}^{c}D_{t}^{q}g(t)=\frac{1}{\Gamma(1-q)} \int_{0}^{t} \frac{g'(s)}{(t-s)^{q}}\,ds,\quad t > 0, 0 < q < 1. $$

As in [16, 17], we state the concept of almost sectorial operators as follows.

Definition 2.3

Let \(-1 < \gamma< 0\) and \(0 < \omega< \frac{\pi}{2}\). By \(\Theta^{\gamma}_{\omega}(X)\) we denote the family of all linear closed operators \(A : D(A) \subset X \rightarrow X\) which satisfy

  1. (1)

    \(\sigma(A)\subset S_{\omega} = \{z \in \mathbb{C}\setminus \{0\}; \vert \arg z\vert \leq\omega\} \cup\{0\}\) and

  2. (2)

    for every \(\omega< \zeta< \pi\) there exists a constant \(C_{\zeta}\) such that

    $$\bigl\Vert R(z;A)\bigr\Vert _{L(X)}\leq C_{\zeta} \vert z \vert ^{\gamma}, \quad\text{for all } z \in\mathbb{C} \setminus S_{\zeta}. $$

A linear operator A will be called an almost sectorial operator on X if \(A \in\Theta^{\gamma}_{\omega}(X)\).

Remark 2.4

Let \(A \in\Theta^{\gamma}_{\omega}(X)\), then the definition implies that \(0 \in\rho(A)\).

We denote the semigroup associated with A by \(W(t)\). For \(t\in S_{\frac{\pi}{2}-\omega}^{0}=\{z\in\mathbb{C}\backslash\{0\}; \vert \arg z\vert <\frac{\pi}{2}-\omega\}\),

$$W(t)=e^{-tz}(A)=\frac{1}{2\pi i} \int_{\Gamma_{\theta}}e^{-tz}R(z;A)\,dz, $$

forms an analytic semigroup of growth order \(1 + \gamma\), here \(\omega< \theta<\frac{\pi}{2}- \vert \arg t\vert \), the integral contour \(\Gamma_{\theta}:= \{\mathbb{R}_{+}e^{i\theta}\} \cup \{\mathbb{R}_{+}e^{-i\theta} \}\) is oriented counter-clockwise [16, 17].

Proposition 2.5

[16, 17]

Let \(A \in\Theta^{\gamma}_{\omega }(X)\) with \(-1 < \gamma< 0\) and \(0 < \omega< \frac{\pi}{2}\). Then

  1. (i)

    There exists a constant \(C_{0}=C_{0}(\gamma)> 0\) such that

    $$\bigl\Vert W(t)\bigr\Vert _{L(X)} \leq C_{0} t^{-\gamma-1},\quad \textit{for all } t > 0. $$
  2. (ii)

    If \(\beta> 1+\gamma\), then \(D(A^{\beta})\subset\Sigma_{T} =\{x\in X; \lim_{t\rightarrow0; t>0}W(t)x=x\}\).

  3. (iii)

    The functional equation \(W(s+t)=W(s)W(t)\) for all \(s, t \in S^{0}_{\frac{\pi}{2}-\omega}\) holds. However, it is not satisfied for \(t=0\) or \(s=0\).

Consider the function of Wright-type (see [17, 18])

$$\begin{aligned} \Psi_{q}(z):= \sum_{n=0}^{\infty} \frac{(-z)^{n}}{n!\Gamma(-q n+1-q)}= \frac{1}{\pi}\sum_{n=1}^{\infty} \frac{(-z)^{n}}{(n-1)!}\Gamma(nq)\sin(n \pi q), \quad z\in\mathbb{C} \end{aligned}$$

with \(0 < q< 1\).

Define operator families \(\{\mathcal{S}_{q}(t)\}|_{t\in S_{\frac{\pi }{2}-\omega}^{0}}\) and \(\{\mathcal{T}_{q}(t)\}|_{t\in S_{\frac{\pi}{2}-\omega}^{0}}\) by

$$\begin{aligned} &\mathcal{S}_{q}(t)x= \int_{0}^{\infty} \Psi_{q} (\sigma)W \bigl( \sigma t^{q} \bigr)x\,d\sigma, \quad t\in S_{\frac{\pi}{2}-\omega}^{0} , x \in X, \\ &\mathcal{T}_{q}(t)x= \int_{0}^{\infty} q \sigma\Psi_{q} (\sigma)W \bigl(\sigma t^{q} \bigr)x\,d\sigma,\quad t\in S_{\frac{\pi}{2}-\omega}^{0} , x\in X. \end{aligned}$$

Theorem 2.6

[18]

For each fixed \(t\in S^{0}_{\frac{\pi}{2}-\omega}\), \(\mathcal{S}_{q}(t)\) and \(\mathcal{T}_{q}(t)\) are linear and bounded operators on X. Moreover, for all \(t > 0\), \(-1<\gamma<0\), \(0< q<1\),

$$\begin{aligned} \begin{aligned} &\bigl\Vert \mathcal{S}_{q}(t)\bigr\Vert \leq M_{1}t^{-q(1+\gamma)}, \\ &\bigl\Vert \mathcal{T}_{q}(t)\bigr\Vert \leq M_{2}t^{-q(1+\gamma)}, \end{aligned} \end{aligned}$$
(2.1)

where \(M_{1}=\frac{C_{0}\Gamma(-\gamma)}{\Gamma(1-q(1+\gamma))}\) and \(M_{2}=\frac{qC_{0}\Gamma(1-\gamma)}{\Gamma(1-q\gamma)}\). Moreover,

$$\begin{aligned} \lim_{t\rightarrow\infty}\bigl\Vert \mathcal{S}_{q}(t) \bigr\Vert =0,\qquad \lim_{t\rightarrow\infty}\bigl\Vert \mathcal{T}_{q}(t) \bigr\Vert =0. \end{aligned}$$
(2.2)

Theorem 2.7

[18], Theorem 3.2

For \(t > 0\), \(\mathcal{S}_{q}(t)\) and \(\mathcal{T}_{q}(t)\) are continuous in the uniform operator topology. Moreover, for every \(\widetilde{r} > 0\), the continuity is uniform on \([\widetilde{r},\infty)\).

Remark 2.8

[18], Theorem 3.4

Let \(\beta>1+\gamma\). Then, for all \(x\in D(A^{\beta})\),

$$\lim_{t\rightarrow0;t>0} \mathcal{S}_{q}(t)x = x. $$

When \(\phi(0) \in D(A^{\beta} )\) with \(\beta>1+\gamma\), we present the definition of mild solution of problem (1.1) as follows.

Definition 2.9

A function \(u\in C([-\delta, +\infty), X)\) satisfying the equation

$$\begin{aligned} u(t)= \textstyle\begin{cases} \phi(t), & t\in[-\delta, 0],\\ {\mathcal{S}_{q}(t)\phi(0)+\int_{0}^{t} (t-s)^{q-1}\mathcal{T}_{q}(t-s)f(s, u_{s})\,ds}, & t>0, \end{cases}\displaystyle \end{aligned}$$
(2.3)

is called a mild solution of problem (1.1).

Remark 2.10

In general, mild solutions to problem (1.1) are assumed to have the same kind of singularity at \(t = 0\) as the operator \(\mathcal{S}_{q}(t)\). When \(\phi(0) \in D(A^{\beta} )\) with \(\beta>1+\gamma\), it follows from Remark 2.8 that the mild solution is continuous at \(t = 0\).

Let us recall the notion of S-asymptotically T-periodic functions which will come into play later on.

Definition 2.11

[9]

A function \(g \in C_{b}(\mathbb{R}_{+}, X)\) is called S-asymptotically T-periodic if there exists \(T > 0\) such that \({\lim_{t\rightarrow\infty} \Vert g(t+T)-g(t)\Vert =0}\). In this case, we say that T is an asymptotic period of g.

Let \(\operatorname{SAP}_{T}(X)\) represent the space formed for all the X-valued S-asymptotically T-periodic functions endowed with the uniform convergence norm denoted by \(\Vert \cdot \Vert _{\infty}\). Then \(\operatorname{SAP}_{T}(X)\) is a Banach space (see Proposition 3.5, [9]). We set \(\operatorname{SAP}_{T, 0}(X)=\{x\in \operatorname{SAP}_{T}(X);x(0)=0\}\). Clearly, \(\operatorname{SAP}_{T, 0}(X)\) is a closed subspace of \(\operatorname{SAP}_{T}(X)\).

Lemma 2.12

Let \(u: [-\delta, +\infty)\rightarrow X\) be a function with \(u_{0} \in C([-\delta, 0], X)\) and \(u|_{[0, +\infty)}\in \operatorname{SAP}_{T}(X)\). Then the function \(t \rightarrow u_{t}\) belongs to \(\operatorname{SAP}_{T}(C([-\delta, 0], X))\).

Proof

Since \(u_{t}\) is continuous on \([-\delta, 0]\) which is compact, there exists \(\overline{\theta}\in[-\delta, 0]\) such that

$$\begin{aligned} \Vert u_{t+T}-u_{t}\Vert _{[-\delta, 0]}=\sup _{-\delta\leq\theta\leq0} \bigl\Vert u(t+T+\theta)-u(t+\theta)\bigr\Vert =\bigl\Vert u(t+T+\overline{\theta})-u(t+\overline {\theta}) \bigr\Vert . \end{aligned}$$

Setting \(\tau=t+\overline{\theta}\), we obtain \(\lim_{t\rightarrow+\infty} \Vert u(t+T+\overline{\theta})-u(t+\overline {\theta})\Vert =\lim_{\tau\rightarrow+\infty} \Vert u(\tau+T)- u(\tau)\Vert =0\). □

3 Main result

In this section we discuss the existence and uniqueness of S-asymptotically T-periodic solutions for problem (1.1).

The function \(f:(0, +\infty) \times C([-\delta, 0], X)\rightarrow X \) satisfies the following conditions:

  1. (H1)

    There exists a function \(\mu(\cdot)\in L^{\frac {1}{p}}(0< p<-q\gamma)\) such that \(\Vert f(t, \varphi)\Vert \leq\mu(t)\), for all \(\varphi\in C([-\delta, 0], X)\).

  2. (H2)

    There exists a function \(\eta:(0, +\infty)\rightarrow(0, +\infty)\) such that

    $$\bigl\Vert f(t, \psi_{1})-f(t, \psi_{2})\bigr\Vert \leq \eta(t)\Vert \psi_{1}-\psi_{2}\Vert ,\quad \text{for all } t> 0, \psi_{1}, \psi_{2}\in C \bigl([-\delta, 0], X \bigr) $$

    and

    $$\begin{aligned} \Lambda:=\sup_{t\geq0} \int_{0}^{t}\frac{\eta (s)}{(t-s)^{1+q\gamma}}\,ds< \frac{1}{M_{2}}. \end{aligned}$$
    (3.1)
  3. (H3)

    \({K:=\sup_{t\geq0}\int_{0}^{t}\frac{\Vert f(s, 0)\Vert }{(t-s)^{1+q\gamma}}\,ds<\infty}\).

  4. (H4)

    There exists a function \(\xi:(0, +\infty)\rightarrow(0, +\infty)\) such that

    $$\lim_{t\rightarrow\infty}\frac{\Vert f(t+T, \varphi)-f(t, \varphi)\Vert }{\xi (t)}=0,\quad \text{for all } \varphi\in C \bigl([-\delta, 0], X \bigr), $$

    and \({\sup_{t\geq0}\int_{0}^{t}\frac{\xi(s)}{(t-s)^{1+q\gamma}}\,ds< \infty}\).

For \(\nu>p\), \(0\leq t_{0}< t\), it follows from the Hölder inequality that

$$\begin{aligned} \int_{t_{0}}^{t} (t-s)^{\nu-1}\mu(s)\,ds\leq \biggl( \int _{t_{0}}^{t}(t-s)^{\frac{\nu-1}{1-p}}\,ds \biggr)^{1-p}\Vert \mu \Vert _{L^{\frac{1}{p}}}= \delta_{\nu}(t-t_{0})^{\nu-p}, \end{aligned}$$
(3.2)

where \({\delta_{\nu}= (\frac{1-p}{\nu-p} )^{1-p}\Vert \mu \Vert _{L^{\frac{1}{p}}}}\).

Theorem 3.1

Assume that (H1)-(H4) hold. Then, for every \(\phi(0) \in D(A^{\beta})\) with \(\beta>1+\gamma\), the problem (1.1) has a unique S-asymptotically T-periodic mild solution.

Proof

For \(\phi\in C([-\delta,0],X)\), we define the function \(y(t)=\phi(t)\) for \(t\in[-\delta, 0]\), \(y(t)=\mathcal{S}_{q}(t)\phi(0)\) for \(t>0\), then \(y\in C([-\delta,\infty), X)\). Set \(u(t)=x(t)+y(t)\), \(t \in[-\delta, +\infty)\). It is obvious that u satisfies (2.3) if and only if x satisfies \(x_{0}=0\) and for \(t> 0\),

$$\begin{aligned} x(t)= { \int_{0}^{t}(t-s)^{q-1}\mathcal{T}_{q}(t-s)f(s, x_{s}+y_{s})\,ds}. \end{aligned}$$

We write

$$\widetilde{C}_{b}(X)= \bigl\{ x\in C_{b} \bigl([-\delta, + \infty ), X\bigr); x|_{t> 0}\in C_{b}(\mathbb{R}_{+}, X), x|_{[-\delta, 0]}=0 \bigr\} , $$

endowed with the norm \(\Vert x\Vert _{\infty}:=\sup_{t> 0}\Vert x(t)\Vert +\Vert x_{0}\Vert _{[-\delta, 0]}=\sup_{t> 0}\Vert x(t)\Vert \).

For \(x\in\widetilde{C}_{b}(X)\), setting \(C_{1}:=\Vert x\Vert _{\infty}+\sup_{t>0 }\Vert \mathcal{S}_{q}(t)\phi(0)\Vert +\Vert \phi \Vert _{[-\delta, 0]}\), we have

$$\begin{aligned} \Vert x_{t}+y_{t}\Vert _{[-\delta, 0]}& \leq \sup_{-\delta\leq\theta \leq0}\bigl\Vert x(t+\theta)\bigr\Vert +\sup _{-\delta\leq\theta\leq0}\bigl\Vert y(t+\theta) \bigr\Vert \\ &\leq \sup_{0< \tau\leq t}\bigl\Vert {x}(\tau)\bigr\Vert +\sup _{0< \tau\leq t}\bigl\Vert \mathcal {S}_{q}(\tau)\phi(0)\bigr\Vert +\Vert \phi \Vert _{[-\delta, 0]}\leq C_{1}, \end{aligned}$$
(3.3)

then

$$\begin{aligned} \bigl\Vert f(t, x_{t}+y_{t})\bigr\Vert \leq\eta(t) \Vert x_{t}+y_{t}\Vert _{[-\delta, 0]}+\bigl\Vert f(t, 0)\bigr\Vert \leq C_{1} \eta(t)+\bigl\Vert f(t, 0)\bigr\Vert . \end{aligned}$$
(3.4)

We consider the operator \(\mathcal{F}\) on \(\widetilde{C}_{b}(X)\) as follows:

$$\begin{aligned} (\mathcal{F}x) (t)= \textstyle\begin{cases} 0, & t\in[-\delta, 0],\\ {\int_{0}^{t}(t-s)^{q-1}\mathcal{T}_{q}(t-s)f(s, x_{s}+y_{s})\,ds}, & t> 0. \end{cases}\displaystyle \end{aligned}$$

We will show initially that \(\mathcal{F}x\in\widetilde{C}_{b}(X)\). Let \(h> 0\), we have

$$\begin{aligned} & \biggl\Vert \int_{0}^{t+h} (t+h-s)^{q-1} \mathcal{T}_{q}(t+h-s)f(s, x_{s}+y_{s})\,ds \\ &\qquad{}- \int_{0}^{t}(t-s)^{q-1} \mathcal{T}_{q}(t-s)f(s, x_{s}+y_{s})\,ds \biggr\Vert \\ &\quad\leq \biggl\Vert \int_{0}^{t} \bigl[(t+h-s)^{q-1}-(t-s)^{q-1} \bigr]\mathcal {T}_{q}(t+h-s)f(s, x_{s}+y_{s}) \,ds \biggr\Vert \\ &\qquad{}+ \biggl\Vert \int_{0}^{t}(t-s)^{q-1} \bigl[ \mathcal{T}_{q}(t+h-s)-\mathcal {T}_{q}(t-s) \bigr]f(s, x_{s}+y_{s})\,ds \biggr\Vert \\ &\qquad{}+ \biggl\Vert \int_{t}^{t+h}(t+h-s)^{q-1} \mathcal{T}_{q}(t+h-s)f(s, x_{s}+y_{s})\,ds \biggr\Vert . \end{aligned}$$

For \(\varepsilon>0\) small enough, we conclude

$$\begin{aligned} &\biggl\Vert \int_{0}^{t} \bigl[(t+h-s)^{q-1}-(t-s)^{q-1} \bigr]\mathcal{T}_{q}(t+h-s)f(s, x_{s}+y_{s})\,ds \biggr\Vert \\ &\quad\leq M_{2} \int_{0}^{t-\varepsilon }\bigl\vert (t+h-s)^{q-1}-(t-s)^{q-1} \bigr\vert (t+h-s)^{-q(1+\gamma)} \mu(s)\,ds \\ &\qquad{}+M_{2} \int_{t-\varepsilon}^{t} \biggl[(t+h-s)^{-1-q\gamma}+ \frac {(t-s)^{q-1}}{(t+h-s)^{q(1+\gamma)}} \biggr]\mu(s)\,ds, \end{aligned}$$

taking \(h\rightarrow0\), \(\varepsilon\rightarrow0\), and using (3.2), the right-hand side of the above inequality tends to zero.

Moreover, by (3.2), we have

$$\begin{aligned} \biggl\Vert \int_{t}^{t+h}(t+h-s)^{q-1} \mathcal{T}_{q}(t+h-s)f(s, x_{s}+y_{s})\,ds \biggr\Vert &\leq M_{2} \int_{t}^{t+h}(t+h-s)^{-q\gamma-1}\mu(s)\,ds \\ &\rightarrow 0,\quad \text{as } h\rightarrow0. \end{aligned}$$

For \(\varepsilon>0\) small enough, noting that (2.1) and (3.2), we obtain

$$\begin{aligned} & \biggl\Vert \int_{0}^{t}(t-s)^{q-1} \bigl[ \mathcal{T}_{q}(t+h-s)-\mathcal {T}_{q}(t-s) \bigr]f(s, x_{s}+y_{s})\,ds \biggr\Vert \\ &\quad\leq \int_{0}^{t-\varepsilon}(t-s)^{q-1}\bigl\Vert \mathcal{T}_{q}(t+h-s)-\mathcal {T}_{q}(t-s)\bigr\Vert _{L(X)} \mu(s) \,ds \\ &\qquad{}+ \int_{t-\varepsilon}^{t}(t-s)^{q-1}\bigl\Vert \mathcal{T}_{q}(t+h-s)-\mathcal {T}_{q}(t-s)\bigr\Vert \mu(s) \,ds \\ &\quad\leq \sup_{s\in[0, t-\varepsilon]}\bigl\Vert \mathcal{T}_{q}(t+h-s)- \mathcal {T}_{q}(t-s)\bigr\Vert _{L(X)}\cdot \int_{0}^{t-\varepsilon}(t-s)^{q-1}\mu(s)\,ds \\ &\qquad{}+ M_{2} \int_{t-\varepsilon}^{t} \biggl(\frac{(t-s)^{q-1}}{(t+h-s)^{q(1+\gamma)}}+ \frac {(t-s)^{q-1}}{(t-s)^{q(1+\gamma)}} \biggr)\mu(s)\,ds. \end{aligned}$$

This, together with Theorem 2.7, shows that the right-hand side tends to zero as \(h \rightarrow0\) and \(\varepsilon\rightarrow0\).

Moreover, from (H2), (H3), (3.1), and (3.4), we have

$$\begin{aligned} \biggl\Vert \int_{0}^{t}\frac{\mathcal{T}_{q}(t-s)f(s, x_{s}+y_{s})}{(t-s)^{1-q}}\,ds \biggr\Vert \leq M_{2} C_{1} \int_{0}^{t}\frac{\eta (s)}{(t-s)^{1+q\gamma}}\,ds+M_{2} K< C_{1}+M_{2}K. \end{aligned}$$

Now, the operator \(\mathcal{F}:\widetilde{C}_{b}(X)\rightarrow \widetilde{C}_{b}(X)\) is well defined. It is clear that the fixed points of \(\mathcal{F}\) are mild solutions to problem (1.1).

Next, we will show that \(\mathcal{F}\) is \(\operatorname{SAP}_{T, 0}(X)\)-valued, where we identify the element \(v\in \operatorname{SAP}_{T, 0}(X)\) with its extension to \([-\delta, +\infty)\) given by \(v_{0}=0\). We will prove for any \(x\in \operatorname{SAP}_{T, 0}(X)\), \(\mathcal{F}x\in \operatorname{SAP}_{T, 0}(X)\). Obviously, (2.2) implies that \(y|_{[0, \infty)}\in \operatorname{SAP}_{T}(X)\), then from Lemma 2.12, the function \(t\rightarrow y_{t}\) belongs to \(\operatorname{SAP}_{T}(C([-\delta, 0], X))\). Now \(x_{t}+y_{t}\in \operatorname{SAP}_{T}(C([-\delta, 0], X))\), this means that, for each \(\varepsilon>0\), there is a positive constant \(L_{1}>0\) such that

$$\begin{aligned} \bigl\Vert (x_{t+T}+y_{t+T})-(x_{t}+y_{t}) \bigr\Vert _{[-\delta, 0]}\leq \varepsilon, \quad\text{for every } t\geq L_{1}. \end{aligned}$$
(3.5)

Moreover, (H4) implies that there is a positive constant \(L_{2}>0\) such that

$$\bigl\Vert f(t+T, x_{t+T}+y_{t+T})-f(t, x_{t+T}+y_{t+T}) \bigr\Vert < \xi(t)\varepsilon, \quad\text{for every } t\geq L_{2}. $$

Then, for \(t>L+T\), \(L:=\max\{ L_{1}, L_{2}\}\), we have

$$\begin{aligned} &\bigl\Vert (\mathcal{F}x) (t+T)-(\mathcal{F}x) (t)\bigr\Vert \\ &\quad=\biggl\Vert \int_{0}^{T}\frac {\mathcal{T}_{q}(t+T-s)}{(t+T-s)^{1-q}}f(s, x_{s}+y_{s})\,ds+ \int _{T}^{t+T}\frac{\mathcal{T}_{q}(t+T-s)}{(t+T-s)^{1-q}}f(s, x_{s}+y_{s})\,ds \\ &\qquad{}- \int_{0}^{t} (t-s)^{q-1} \mathcal{T}_{q}(t-s)f(s, x_{s}+y_{s})\,ds\biggr\Vert \\ &\quad\leq M_{2} \biggl[ \int_{0}^{T}(t+T-s)^{-1-q\gamma} \bigl(C_{1} \eta(s)+\bigl\Vert f(s, 0)\bigr\Vert \bigr)\,ds \biggr] \\ &\qquad{}+ \int_{0}^{t}(t-s)^{q-1}\bigl\Vert \mathcal{T}_{q}(t-s)\bigr\Vert \bigl\Vert f(s+T, x_{s+T}+y_{s+T})-f(s, x_{s+T}+y_{s+T}) \bigr\Vert \,ds \\ &\qquad{}+M_{2} \int_{0}^{t} (t-s)^{-1-q\gamma}\bigl\Vert f(s, x_{s+T}+y_{s+T})-f(s, x_{s}+y_{s})\bigr\Vert \,ds \\ &\quad =I_{1}(t)+I_{2}(t)+I_{3}(t). \end{aligned}$$

Noting that \({t+T-s\geq\frac{t+T}{T}(T-s)}\), we have

$$\begin{aligned} \int_{0}^{T}\frac{C_{1} \eta(s)+\Vert f(s, 0)\Vert }{(t+T-s)^{1+q\gamma}}\,ds \leq& \biggl( \frac{T}{t+T} \biggr)^{1+q\gamma} \int_{0}^{T}\frac{C_{1} \eta(s)+\Vert f(s, 0)\Vert }{(T-s)^{1+q\gamma}}\,ds, \end{aligned}$$

which implies \(I_{1}(t)\rightarrow0\) as \(t\rightarrow\infty\). From (3.4),

$$\bigl\Vert f(s+T, x_{s+T}+y_{s+T})-f(s, x_{s+T}+y_{s+T}) \bigr\Vert \leq C_{1} \bigl[\eta (s+T)+\eta(s) \bigr]+\bigl\Vert f(s+T, 0)\bigr\Vert +\bigl\Vert f(s, 0) \bigr\Vert . $$

Denoting \({\sup_{t\geq0}\int_{0}^{t}\frac{\xi(s)}{(t-s)^{1+q\gamma }}\,ds:=\widetilde{M}}\) and noting that (H2), (H3), (2.2), and \({t-s\geq\frac{t}{L}(L-s)}\), we have

$$\begin{aligned} I_{2}(t) \leq& M_{2} \int_{0}^{L}\frac{C_{1}\eta(s+T)+\Vert f(s+T, 0)\Vert +C_{1} \eta(s)+\Vert f(s, 0)\Vert }{(t-s)^{1+q\gamma}}\,ds \\ &{}+M_{2}\varepsilon \int_{L}^{t}(t-s)^{-1-q \gamma}\xi(s)\,ds \\ \leq&M_{2} \biggl[ \int_{T}^{L+T}\frac{C_{1}\eta(s)+\Vert f(s, 0)\Vert }{(t+T-s)^{1+q\gamma}}\,ds + \int_{0}^{L}\frac{C_{1}\eta(s)+\Vert f(s, 0)\Vert }{(t-s)^{1+q\gamma}}\,ds \biggr]+M_{2}\widetilde{M} \varepsilon \\ \leq& M_{2} \biggl(\frac{L+T}{t+T} \biggr)^{1+q\gamma} \biggl[ \int_{0}^{L+T} \frac{(C_{1}\eta(s)+\Vert f(s, 0)\Vert )\,ds}{(L+T-s)^{1+q\gamma}} \biggr] \\ &{}+M_{2} \biggl(\frac{L}{t} \biggr)^{1+q\gamma} \biggl[ \int_{0}^{L} \frac{(C_{1} \eta(s)+\Vert f(s, 0)\Vert )\,ds}{(L-s)^{1+q\gamma}} \biggr]+M_{2}\widetilde{M} \varepsilon \\ \leq& 2M_{2} \biggl(\frac{L+T}{t+T} \biggr)^{1+q\gamma}[C_{1} \Lambda +K]+M_{2}\widetilde{M} \varepsilon. \end{aligned}$$

From (H2), (3.3), and (3.5), it follows that

$$\begin{aligned} I_{3}(t) \leq& 2M_{2}C_{1} \int_{0}^{L} \frac{\eta(s)}{(t-s)^{1+q\gamma }}\,ds+M_{2} \varepsilon \int_{L}^{t} \frac{\eta(s)}{(t-s)^{1+q\gamma}}\,ds \\ \leq&M_{2}\Lambda \biggl[2C_{1} \biggl(\frac{L}{t} \biggr)^{1+q\gamma }+\varepsilon \biggr]. \end{aligned}$$

Now we can see \(\Vert (\mathcal{F}x)(t+T)-(\mathcal{F}x)(t)\Vert \rightarrow0\) as \(t\rightarrow\infty\). As a result, \(\mathcal{F}(\operatorname{SAP}_{T, 0}(X))\subseteq \operatorname{SAP}_{T, 0}(X)\).

For \(x, \widetilde{x}\in \operatorname{SAP}_{T, 0}(X)\), we have

$$\begin{aligned} \bigl\Vert (\mathcal{F}x) (t)-(\mathcal{F}\widetilde{x}) (t)\bigr\Vert \leq& M_{2} \int_{0}^{t}(t-s)^{-1-q\gamma}\eta(s) \Vert x_{s}-\widetilde{x}_{s}\Vert _{[-\delta, 0]}\,ds\leq M_{2}\Lambda \Vert x-\widetilde{x}\Vert _{\infty}, \end{aligned}$$

then \({ \Vert \mathcal{F}x-\mathcal{F}\widetilde{x}\Vert _{\infty}\leq M_{2}\Lambda \Vert x-\widetilde{x}\Vert _{\infty}}\). Then \(\mathcal{F}\) is a contraction mapping, the proof now can be finished by using the contraction mapping principle. □

Example 3.2

Let Ω be a bounded domain in \(\mathbb{R}^{N}\ (N\geq1)\) with boundary ∂Ω of class \(C^{4}\). Let \(X=C^{\frac {4}{5}}(\overline{\Omega})\),

$$\widetilde{A}=\Delta,\quad D(\widetilde{A})= \bigl\{ u\in C^{\frac{14}{5}}(\overline { \Omega});u|_{\partial\Omega}=0 \bigr\} . $$

It follows from Example 2.3 [17] that there exist \(\varsigma, \omega>0\), such that

$$\widetilde{A}+\varsigma\in\Theta^{-\frac{3}{5}}_{\frac{\pi}{2}-\omega} \bigl(C^{\frac{4}{5}}(\overline{\Omega}) \bigr), $$

then \(\widetilde{A}+\varsigma\) is an almost sectorial operator and generates a semigroup \(\{W(t)\}_{t\geq0}\) with \(\Vert W(t)\Vert \leq C_{0} t^{-\gamma-1}(\gamma=-\frac{3}{5}, C_{0}>0)\).

We consider the following fractional differential problem:

$$\begin{aligned} \textstyle\begin{cases} {^{c}D_{t}^{\frac{5}{6}} v(t, x)=\Delta v(t, x)+2v(t, x)+ \frac{\alpha\cos v(t+\tau, x)}{(t+t^{4})^{\frac{1}{4}}}+\frac{\sin2\pi t}{\sqrt{t+1}}},\quad t>0, x\in \Omega,\\ v|_{\partial\Omega}=0,\\ v(t, x)=\phi(t, x),\quad t\in[-1, 0], \end{cases}\displaystyle \end{aligned}$$
(3.6)

where \(\alpha\in\mathbb{R}\) and \(\phi:[-1, 0]\times\Omega\rightarrow X\) is a continuous function.

Problem (3.6) can be written in the abstract form as follows:

$$\begin{aligned} &^{c} D_{t}^{q} v(t)= Av(t)+F(t, v_{t}), \quad t> 0, \\ &v(t)=\phi(t), \quad t\in[-1, 0], \end{aligned}$$

where \(q=\frac{5}{6}, Av=(\widetilde{A}+2)v, {F(t, v_{t})=\frac{\alpha \cos v_{t}}{(t+t^{4})^{\frac{1}{4}}}+\frac{\sin2\pi t}{\sqrt{t+1}}}\).

For \(\varphi, \widetilde{\varphi} \in C([-1, 0], X)\), we can easily see

$$\begin{aligned} &\bigl\Vert F(t, \varphi)\bigr\Vert \leq \frac{\vert \alpha \vert }{(t+t^{4})^{\frac{1}{4}}}+ \frac {1}{\sqrt{t+1}}:=\mu(t), \\ &\bigl\Vert F(t, \varphi)-F(t, \widetilde{\varphi})\bigr\Vert \leq \frac{\vert \alpha \vert }{(t+t^{4})^{\frac{1}{4}}} \Vert \varphi- \widetilde{\varphi} \Vert :=\eta(t)\Vert \varphi-\widetilde{\varphi} \Vert . \end{aligned}$$

Noting that

$$\begin{aligned} &\int_{0}^{1} \biggl(\frac{1}{t+t^{4}} \biggr)^{\frac{3}{4}}\,dt\leq \int_{0}^{1} \biggl(\frac{1}{t} \biggr)^{\frac{3}{4}}\,dt=4, \\ &\int_{1}^{+\infty} \biggl(\frac{1}{t+t^{4}} \biggr)^{\frac{3}{4}}\,dt\leq \int _{1}^{+\infty} \biggl(\frac{1}{t^{4}} \biggr)^{\frac{3}{4}}\,dt= \int_{1}^{+\infty }\frac{1}{t^{3}}\,dt= \frac{1}{2}, \\ &\int_{0}^{+\infty} \biggl(\frac{1}{1+t} \biggr)^{\frac{3}{2}}\,dt=2, \end{aligned}$$

we can obtain \(\mu(t)\in L^{\frac{1}{p}}(p=\frac{1}{3})\).

Noting that \(-1-q\gamma=-\frac{1}{2}\), \({(t+t^{4})^{-\frac{1}{4}}\leq t^{-\frac{1}{4}}}\), \({\int_{0}^{t} (t-s)^{-\frac{1}{2}}s^{-\frac {1}{2}}\,ds=\pi}\), we have

$$\begin{aligned} \Lambda :=\sup_{t\geq0} \int_{0}^{t} (t-s)^{-\frac{1}{2}}\eta(s)\,ds &= \vert \alpha \vert \sup_{t\geq0} \int_{0}^{t} (t-s)^{-\frac{1}{2}}s^{-\frac {1}{4}} \cdot \bigl(1+s^{3} \bigr)^{-\frac{1}{4}}\,ds \\ &= \vert \alpha \vert \sup_{t\geq0} \int_{0}^{t} (t-s)^{-\frac{1}{2}}s^{-\frac {1}{2}} \cdot s^{\frac{1}{4}} \bigl(1+s^{3} \bigr)^{-\frac{1}{4}}\,ds \\ &\leq \vert \alpha \vert \pi. \end{aligned}$$
(3.7)

Moreover, \({F(t, 0)=\frac{\alpha}{(t+t^{4})^{\frac{1}{4}}} +\frac{\sin 2\pi t}{\sqrt{t+1}}}\), noting that (3.7) and

$$\int_{0}^{t}(t-s)^{-\frac{1}{2}}\frac{\vert \sin2\pi s\vert }{\sqrt{s+1}} \,ds \leq \int _{0}^{t}(t-s)^{-\frac{1}{2}}s^{-\frac{1}{2}} \,ds= \pi, $$

we get

$$\sup_{t\geq0} \int_{0}^{t}(t-s)^{-\frac{1}{2}} \bigl\Vert F(s, 0) \bigr\Vert \,ds< + \infty. $$

We take \(\xi(t)=t^{-\frac{1}{2}}\) then

$$\begin{aligned} \frac{\Vert F(t+1, \varphi)-F(t, \varphi)\Vert }{t^{-\frac{1}{2}}} \leq{}&\vert \alpha \vert t^{\frac{1}{2}} \biggl[ \frac{1}{(t+t^{4})^{\frac {1}{4}}}+ \frac{1}{[(t+1)+(t+1)^{4}]^{\frac{1}{4}}} \biggr] \\ &{}+t^{\frac{1}{2}} \biggl[\frac{1}{\sqrt{t+1}}-\frac{1}{\sqrt{t+2}} \biggr] \\ \rightarrow{}& 0,\quad \text{as } t\rightarrow+\infty, \end{aligned}$$

and \({\sup_{t\geq0}\int_{0}^{t}(t-s)^{-\frac{1}{2}}s^{-\frac {1}{2}}\,ds=\pi< \infty}\).

Noting that \({M_{2}=\frac{qC_{0}\Gamma(1-\gamma)}{\Gamma(1-q\gamma)}=\frac {5C_{0}\Gamma(\frac{8}{5})}{6\Gamma(\frac{3}{2})}}\) and (3.7), in association with Theorem 3.1, if \({\vert \alpha \vert <\frac{6\Gamma(\frac{3}{2})}{5\pi C_{0} \Gamma(\frac{8}{5})}}\), then the problem (3.6) has a unique S-asymptotically 1-periodic mild solution for \(\phi(0, x)\in D((A+2)^{\frac{3}{5}})\).