We make the following assumptions:

\((H_{1})\) The function \(f: J\times X_{\alpha}^{n+1}\rightarrow X\) is continuous and there exist positive constants \(\beta_{0}, \beta_{1}, \ldots, \beta_{n}\) and \(K\geq0\) such that

$$\bigl\Vert f(t,\nu_{0},\nu_{1},\ldots,\nu_{n}) \bigr\Vert \leq\sum_{i=0}^{n}\beta _{i} \Vert \nu_{i} \Vert _{\alpha}+K, \quad t\in J, (\nu_{0},\nu_{1},\ldots,\nu _{n})\in X_{\alpha}^{n+1}. $$

\((H_{2})\) The function \(g: C([-r,T], X_{\alpha})\rightarrow X_{\alpha }\) is continuous and there exists a constant \(L\geq M\) such that

$$\bigl\Vert g(x)-g(y) \bigr\Vert _{\alpha}\leq\frac{\|x-y\|_{C}}{L+\|x-y\|_{C}}, \quad x, y\in C\bigl([-r,T], X_{\alpha}\bigr). $$

### Remark 2

The condition \((H_{1})\) can be replaced by the condition

\((H_{1})'\) The function \(f: J\times X_{\alpha}^{n+1}\rightarrow X\) is continuous and there exist functions \(\overline{\beta}_{i}\in L^{1}(J, {\mathbb{R}} ^{+}), i=0,1,\ldots,n\) and nondecreasing function \(\Phi: [0,+\infty )\rightarrow(0,+\infty)\) such that

$$\bigl\Vert f\bigl(t,x(t),x(t-\tau_{1}),\ldots,x(t- \tau_{n})\bigr) \bigr\Vert \leq\sum_{i=0}^{n} \overline{\beta}_{i}(t)\Phi\bigl( \bigl\Vert x(t-\tau_{i}) \bigr\Vert _{\alpha}\bigr), $$

for any \(t\in J, x\in C([-r,T],X_{\alpha})\), where \(\tau_{0}=0\).

By \((H_{1})'\), we have

$$\bigl\Vert f\bigl(t,x(t),x(t-\tau_{1}),\ldots,x(t- \tau_{n})\bigr) \bigr\Vert \leq\sum_{i=0}^{n} \Vert \overline{\beta}_{i} \Vert _{L^{1}}\Phi\bigl( \bigl\Vert x(t-\tau_{i}) \bigr\Vert _{\alpha}\bigr). $$

Let \(\Phi(r)=r\) for some \(r>0\). Then \((H_{1})'\Rightarrow(H_{1})\) with \(\beta_{i}=\|\overline{\beta}_{i}\|_{L^{1}}\) and \(K=0\). Since \((H_{1})'\) would not be an essential generalization, we only consider \((H_{1})\) in the following.

Define an operator \(Q_{1}: C([-r,T], X_{\alpha})\rightarrow C([-r,T], X_{\alpha})\) by

$$ (Q_{1}x) (t)= \textstyle\begin{cases} \varphi(t)-g(x), & t\in[-r,0],\\ U(t)(\varphi(0)-g(x)), & t\in J. \end{cases} $$

(3.1)

### Lemma 8

*If the assumption*
\((H_{2})\)
*holds*, \(Q_{1}\)
*is a nonlinear contraction*.

### Proof

Let \(x,y\in C([-r,T], X_{\alpha})\). For \(t\in[-r,0]\), we have

$$\bigl\Vert (Q_{1}x) (t)-(Q_{1}y) (t) \bigr\Vert _{\alpha}= \bigl\Vert g(x)-g(y) \bigr\Vert _{\alpha}\leq \frac{\| x-y\|_{C}}{L+\|x-y\|_{C}}\leq\frac{M\|x-y\|_{C}}{L+\|x-y\|_{C}}. $$

For \(t\in[0,T]\), we have

$$\begin{aligned} \bigl\Vert (Q_{1}x) (t)-(Q_{1}y) (t) \bigr\Vert _{\alpha} =& \bigl\Vert U(t) \bigl(\varphi (0)-g(x)\bigr)-U(t) \bigl( \varphi(0)-g(y)\bigr) \bigr\Vert _{\alpha} \\ \leq& M \bigl\Vert g(x)-g(y) \bigr\Vert _{\alpha}\leq \frac{M\|x-y\|_{C}}{L+\|x-y\|_{C}}. \end{aligned}$$

This implies that

$$\Vert Q_{1}x-Q_{1}y \Vert _{C}\leq \frac{M\|x-y\|_{C}}{L+\|x-y\|_{C}}. $$

Let \(\phi(r)=\frac{Mr}{L+r}\). Then \(\phi: {\mathbb{R}}^{+}\rightarrow {\mathbb{R}}^{+}\) is continuous and nondecreasing and \(\phi(r)< r\) for \(r>0\). Hence \(Q_{1}\) is a nonlinear contraction in \(C([-r,T], X_{\alpha})\). This completes the proof. □

For any \(\epsilon>0\) and \(h_{1}\in X_{\alpha}\), define a control \(u(t):=u(t;x)\) by

$$\begin{aligned} u(t;x) =& B^{*}V^{*}(T-t)R\bigl(\epsilon,\Gamma_{0}^{T}\bigr) \biggl[h_{1}-U(T) \bigl(\varphi (0)-g(x)\bigr)\\ &{}- \int_{0}^{T}(T-s)^{q-1}V(T-s)F(x) (s)\,ds \biggr]. \end{aligned}$$

Then, from assumptions \((H_{1})\) and \((H_{2})\), we have

$$ \bigl\Vert Bu(t;x) \bigr\Vert _{\alpha}\leq L_{u}, $$

(3.2)

where

$$\begin{aligned} L_{u} =&\frac{1}{\epsilon}L_{B} \bigl\Vert A^{-\alpha} \bigr\Vert \bigl[ \Vert h_{1} \Vert _{\alpha}+M \Vert \varphi \Vert _{C[-r,0]}+M\bigl(1+ \bigl\Vert g(0) \bigr\Vert _{\alpha}\bigr)\bigr] \\ &{}+\frac{L_{B}MKT^{q}}{\epsilon\Gamma(q+1)}+\frac{L_{B}M}{\epsilon \Gamma(q)}\sum_{i=0}^{n} \int_{0}^{T}(T-s)^{q-1}\beta_{i} \bigl\Vert x(s-\tau_{i}) \bigr\Vert _{\alpha}\,ds, \\ L_{B} =& \Vert B \Vert _{\alpha}\sup_{t\in J} \bigl\Vert B^{*}V^{*}(T-t) \bigr\Vert . \end{aligned}$$

Define an operator \(Q_{2}: C([-r,T], X_{\alpha})\rightarrow C([-r,T], X_{\alpha})\) by

$$ (Q_{2}x) (t)= \textstyle\begin{cases} 0, & t\in[-r,0],\\ \int_{0}^{t}(t-s)^{q-1}V(t-s)[F(x)(s)+Bu(s;x)]\,ds, & t\in J. \end{cases} $$

(3.3)

### Lemma 9

*If the assumptions*
\((H_{1})\)
*and*
\((H_{2})\)
*hold*, \(Q_{2}\)
*is completely continuous*.

### Proof

By the assumptions \((H_{1})\) and \((H_{2})\), it is easy to prove that \(Q_{2}: C([-r,T], X_{\alpha})\rightarrow C([-r,T], X_{\alpha})\) is continuous. So, it remains to prove that \(Q_{2}\) is a compact operator on \(C([-r,T], X_{\alpha})\). The case \(t\leq0\) is trivial. Thus, let \(t\in (0,T]\) be fixed. For each \(\delta\in(0,t), \rho>0\) and \(x\in B_{r}:=\{x\in C([-r,T], X_{\alpha}): \|x\|_{C}\leq r\}\), \(r>0\), we define \(Q_{2}^{\delta,\rho}\) by

$$\begin{aligned} \bigl(Q_{2}^{\delta,\rho}x\bigr) (t) =& \int_{0}^{t-\delta}(t-s)^{q-1} \int _{\rho}^{\infty}q\theta\eta_{q}(\theta)S \bigl((t-s)^{q}\theta \bigr)\bigl[F(x) (s)+Bu(s;x)\bigr]\,d\theta \,ds \\ =&S\bigl(\delta^{q}\rho\bigr) \int_{0}^{t-\delta}(t-s)^{q-1} \int_{\rho }^{\infty}q\theta\eta_{q}(\theta)S \bigl((t-s)^{q}\theta-\delta^{q}\rho \bigr)\\ &{}\times\bigl[F(x) (s)+Bu(s;x)\bigr]\,d\theta \,ds. \end{aligned}$$

Then the set \(\{(Q_{2}^{\delta,\rho}x)(t): x\in B_{r}\}\) is relatively compact in \(X_{\alpha}\) because of Lemma 3. By \((H_{1})\), \((H_{2})\) and (3.2), we have

$$\begin{aligned}& \bigl\Vert (Q_{2}x) (t)-\bigl(Q_{2}^{\delta,\rho}x \bigr) (t) \bigr\Vert _{\alpha} \\& \quad\leq \biggl\Vert \int_{0}^{t}(t-s)^{q-1} \int_{0}^{\rho}q\theta\eta _{q}(\theta)S \bigl((t-s)^{q}\theta\bigr) \bigl(F(x) (s)+Bu(s;x)\bigr)\,d\theta \,ds \biggr\Vert _{\alpha } \\& \qquad{}+ \biggl\Vert \int_{t-\delta}^{t}(t-s)^{q-1} \int_{\rho}^{\infty}q\theta \eta_{q}(\theta)S \bigl((t-s)^{q}\theta\bigr) \bigl(F(x) (s)+Bu(s;x)\bigr)\,d\theta \,ds \biggr\Vert _{\alpha} \\& \quad\leq\frac{M_{\alpha}T^{q(1-\alpha)}}{1-\alpha}\Biggl(\sum_{i=0}^{n} \beta_{i}r+K\Biggr) \int_{0}^{\rho}\theta^{1-\alpha}\eta _{q}(\theta)\,d\theta \\& \qquad{}+MT^{q}L_{u} \int_{0}^{\rho}\theta\eta_{q}(\theta)\,d \theta \\& \qquad{}+C_{\alpha}\Biggl(\sum_{i=0}^{n} \beta_{i}r+K\Biggr) \int_{t-\delta }^{t}(t-s)^{q(1-\alpha)-1}\,ds \\& \qquad{}+\frac{ML_{u}\delta^{q}}{\Gamma(q+1)}. \end{aligned}$$

This implies that the set \(\{(Q_{2}x)(t): x\in B_{r}\}\) is relatively compact in \(X_{\alpha}\) for all \(t\in(0, T]\). Hence, we obtain the relative compactness of \((Q_{2}B_{r})(t)\) in \(X_{\alpha}\) for all \(t\in[-r, T]\). We further show that the operator \(Q_{2}\) is equicontinuous in \(C([-r,T], X_{\alpha})\). For \(x\in C([-r,T], X_{\alpha})\), if \(\nu\in[0,T)\), we have

$$\begin{aligned}& \bigl\Vert (Q_{2}x) (\nu)-(Q_{2}x) (0) \bigr\Vert _{\alpha} \\& \quad\leq C_{\alpha} \int_{0}^{\nu}(\nu-s)^{q(1-\alpha)-1}\sum _{i=0}^{n}\beta_{i} \bigl\Vert x(s- \tau_{i}) \bigr\Vert _{\alpha}\,ds \\& \qquad{}+\frac{ML_{u}\nu^{q}}{\Gamma(q+1)}+\frac{C_{\alpha}K\nu ^{q(1-\alpha)}}{q(1-\alpha)} \\& \quad\rightarrow 0 \end{aligned}$$

as \(\nu\rightarrow0\). Hence, it is only necessary to consider the case \(t>0\). For \(0< t_{1}< t_{2}\leq T\), denote

$$\begin{aligned}& I_{1} = \biggl\Vert \int _{0}^{t_{1}}(t_{1}-s)^{q-1} \bigl[V(t_{2}-s)-V(t_{1}-s)\bigr] \bigl[F(x) (s)+Bu(s;x) \bigr]\,ds \biggr\Vert _{\alpha}, \\& I_{2} = \biggl\Vert \int _{0}^{t_{1}}\bigl[(t_{2}-s)^{q-1}-(t_{1}-s)^{q-1} \bigr]V(t_{2}-s)\bigl[F(x) (s)+Bu(s;x)\bigr]\,ds \biggr\Vert _{\alpha}, \\& I_{3}= \biggl\Vert \int _{t_{1}}^{t_{2}}(t_{2}-s)^{q-1}V(t_{2}-s) \bigl[F(x) (s)+Bu(s;x)\bigr]\,ds \biggr\Vert _{\alpha}. \end{aligned}$$

By assumptions \((H_{1}), (H_{2})\) and (3.2), we can find

$$\begin{aligned} I_{2} \leq & C_{\alpha} \int _{0}^{t_{1}} \bigl\vert (t_{2}-s)^{q-1}-(t_{1}-s)^{q-1} \bigr\vert (t_{2}-s)^{-q\alpha} \bigl\Vert F(x) (s) \bigr\Vert \,ds \\ &{}+\frac{ML_{u}[t_{2}^{q}-t_{1}^{q}-(t_{2}-t_{1})^{q}]}{\Gamma(q+1)}, \\ I_{3} \leq & C_{\alpha} \int_{t_{1}}^{t_{2}}(t_{2}-s)^{q(1-\alpha)-1} \bigl\Vert F(x) (s) \bigr\Vert \,ds \\ &{}+\frac{ML_{u}(t_{2}-t_{1})^{q}}{\Gamma(q+1)}. \end{aligned}$$

This implies that \(I_{i}\rightarrow0, i=2,3\) as \(t_{2}-t_{1}\rightarrow0\). For \(t_{1}>0\) and \(\eta\in(0,t_{1})\) small enough, we have

$$\begin{aligned} I_{1} \leq& \int_{0}^{t_{1}-\eta}(t_{1}-s)^{q-1} \bigl\Vert \bigl[V(t_{2}-s)-V(t_{1}-s)\bigr] \bigl[F(x) (s)+Bu(s;x)\bigr] \bigr\Vert _{\alpha}\,ds \\ &{}+ \int_{t_{1}-\eta}^{t_{1}}(t_{1}-s)^{q-1} \bigl\Vert \bigl[V(t_{2}-s)-V(t_{1}-s)\bigr] \bigl[F(x) (s)+Bu(s;x)\bigr] \bigr\Vert _{\alpha}\,ds \\ \leq& \int_{0}^{t_{1}-\eta}(t_{1}-s)^{q-1} \bigl\Vert F(x) (s) \bigr\Vert \,ds\sup_{s\in [0,t_{1}-\eta]} \bigl\Vert V(t_{2}-s)-V(t_{1}-s) \bigr\Vert _{\alpha} \\ &{}+\frac{L_{u}(t_{1}^{q}-\eta^{q})}{q}\sup_{s\in[0,t_{1}-\eta]} \bigl\Vert V(t_{2}-s)-V(t_{1}-s) \bigr\Vert _{\alpha} \\ &{}+C_{\alpha} \int_{t_{1}-\eta}^{t_{1}}(t_{1}-s)^{q-1}(t_{2}-s)^{-q\alpha} \bigl\Vert F(x) (s) \bigr\Vert \,ds \\ &{}+C_{\alpha} \int_{t_{1}-\eta}^{t_{1}}(t_{1}-s)^{q(1-\alpha)-1} \bigl\Vert F(x) (s) \bigr\Vert \,ds \\ &{}+\frac{2ML_{u}\eta^{q}}{\Gamma(q+1)}. \end{aligned}$$

Since compact semigroup is equicontinuous semigroup, it follows that \(I_{1}\rightarrow0\) as \(t_{2}-t_{1}\rightarrow0\) and \(\eta\rightarrow0\). Therefore, from the inequality

$$\bigl\Vert (Q_{2}x) (t_{2})-(Q_{2}x) (t_{1}) \bigr\Vert _{\alpha}\leq I_{1}+I_{2}+I_{3}, $$

we see that the operator \(Q_{2}\) is equicontinuous in \(C([-r,T], X_{\alpha})\). Hence, by the Ascoli-Arzela theorem, \(Q_{2}\) is a compact operator in \(C([-r,T], X_{\alpha})\). This completes the proof. □

### Theorem 1

*Assume that the conditions*
\((H_{1})\)
*and*
\((H_{2})\)
*hold*. *Then the fractional nonlocal control system* (1.1) *has at least one mild solution*.

### Proof

Define two operators \(Q_{1}, Q_{2}: C([-r,T], X_{\alpha })\rightarrow C([-r,T], X_{\alpha})\) as in (3.1) and (3.3). By Lemma 8 and 9, it follows that all the conditions of Lemma 6 are satisfied and a direct application of Lemma 6 shows that either the conclusion (i) or the conclusion (ii) holds. We next show that the conclusion (ii) is not possible. Equivalently, we prove that the set \(\Sigma:=\{x\in C([-r,T], X_{\alpha}): \lambda(Q_{1}x+Q_{2}x)=x, 0<\lambda<1\}\) is bounded.

Let \(x\in C([-r,T], X_{\alpha})\) satisfy the operator equation \(x=\lambda(Q_{1}x+Q_{2}x)\) for some \(\lambda\in(0,1)\). Then, for any \(t\in[-r, 0]\), by assumption \((H_{2})\), we have

$$\bigl\Vert x(t) \bigr\Vert _{\alpha}\leq \Vert \varphi \Vert _{C[-r,0]}+ \bigl\Vert g(x) \bigr\Vert _{\alpha}\leq \Vert \varphi \Vert _{C[-r,0]}+1+ \bigl\Vert g(0) \bigr\Vert _{\alpha} \triangleq\overline{M}_{1}. $$

For \(t\geq0\), by assumptions \((H_{1})\) and \((H_{2})\), we have

$$\begin{aligned} \bigl\Vert x(t) \bigr\Vert _{\alpha} \leq& \bigl\Vert U(t) \bigl( \varphi(0)-g(x)\bigr) \bigr\Vert _{\alpha}+ \biggl\Vert \int _{0}^{t}(t-s)^{q-1}V(t-s)F(x) (s)\,ds \biggr\Vert _{\alpha} \\ &{}+ \biggl\Vert \int_{0}^{t}(t-s)^{q-1}V(t-s)Bu(s;x)\,ds \biggr\Vert _{\alpha} \\ \leq & C+C_{\alpha}\sum_{i=0}^{n} \int _{0}^{t}(t-s)^{q(1-\alpha)-1} \beta_{i} \bigl\Vert x(s-\tau_{i}) \bigr\Vert _{\alpha}\,ds, \end{aligned}$$

where \(\tau_{0}=0\) and \(C=M\|\varphi\|_{C[-r,0]}+M(1+\|g(0)\|_{\alpha })+\frac{KC_{\alpha}T^{q(1-\alpha)}}{q(1-\alpha)}+\frac {ML_{u}T^{q}}{\Gamma(q+1)}\).

Let

$$\psi(t)=\max_{s\in[-r,t]} \bigl\Vert x(s) \bigr\Vert _{\alpha},\quad t\in[-r,T]. $$

Then \(\psi\in C([-r, T], {\mathbb{R}}^{+})\) and \(\|x(t)\|_{\alpha}\leq \psi(t)\) for \(t\in[-r, T]\). For every \(t\geq0\), by the definition of *ψ*, there exists \(\theta_{t}\in[-r, t]\) such that \(\psi(t)=\|x(\theta_{t})\| _{\alpha}\).

If \(-r\leq\theta_{t}\leq0\), we have

$$\begin{aligned} \psi(t) =& \bigl\Vert x(\theta_{t}) \bigr\Vert _{\alpha}\leq \bigl\Vert \varphi(\theta_{t}) \bigr\Vert _{\alpha}+1+ \bigl\Vert g(0) \bigr\Vert _{\alpha} \\ \leq & M \Vert \varphi \Vert _{C[-r,0]}+M \bigl(1+ \bigl\Vert g(0) \bigr\Vert _{\alpha}\bigr) \\ \leq & C+C_{\alpha}\sum_{i=0}^{n} \beta_{i} \int _{0}^{t}(t-s)^{q(1-\alpha)-1}\psi(s)\,ds. \end{aligned}$$

If \(\theta_{t}>0\), we have

$$\begin{aligned} \psi(t) =& \bigl\Vert x(\theta_{t}) \bigr\Vert _{\alpha}\leq C+C_{\alpha} \int _{0}^{\theta_{t}}(\theta_{t}-s)^{q(1-\alpha)-1} \sum_{i=0}^{n}\beta_{i} \bigl\Vert x(s-\tau_{i}) \bigr\Vert _{\alpha}\,ds \\ \leq & C+C_{\alpha}\sum_{i=0}^{n} \beta_{i} \int_{0}^{\theta _{t}}(\theta_{t}-s)^{q(1-\alpha)-1} \psi(s)\,ds \\ \leq& C+C_{\alpha}\sum_{i=0}^{n} \beta_{i} \int _{0}^{t}(t-s)^{q(1-\alpha)-1}\psi(s)\,ds. \end{aligned}$$

Using the well-known singular version of Gronwall inequality [12], we can deduce that there exists a constant \(\overline{M}_{2}>0\) such that \(\psi(t)\leq\overline{M}_{2}\). Thus, for any \(t\geq0\), we have

$$\bigl\Vert x(t) \bigr\Vert _{\alpha}\leq\psi(t)\leq \overline{M}_{2}. $$

Consequently,

$$\|x\|_{C}=\max_{t\in[-r, T]} \bigl\Vert x(t) \bigr\Vert _{\alpha}\leq\overline {M}_{1}+\overline{M}_{2} \triangleq\overline{M}. $$

This implies that the set Σ is bounded. Therefore, by Lemma 6, the operator equation \(x=Q_{1}x+Q_{2}x\) has at least one fixed point which is the mild solution of the fractional control system (1.1) on \(C([-r,T], X_{\alpha})\). This completes the proof. □

### Remark 3

Even if \(g(x)\equiv0\) and without control *u* in the fractional nonlocal control system (1.1), Theorem 1 is still new.

The condition \((H_{2})\) can be replaced by the following condition:

\((H_{2})'\) The function \(g: C([-r,T], X_{\alpha})\rightarrow X_{\alpha }\) is Lipschitz continuous with constant \(L_{1}\in(0,\frac{1}{M})\), that is, for any \(x,y\in C([-r,T], X_{\alpha})\), we have

$$\bigl\Vert g(x)-g(y) \bigr\Vert _{\alpha}\leq L_{1} \Vert x-y \Vert _{C}. $$

### Theorem 2

*Let the conditions*
\((H_{1})\)
*and*
\((H_{2})'\)
*hold*. *Then the fractional nonlocal control system* (1.1) *has at least one mild solution*.

### Proof

By the condition \((H_{2})'\), similar to the proof as in Lemma 8, we obtain

$$\Vert Q_{1}x-Q_{1}y \Vert _{C}\leq ML_{1} \Vert x-y \Vert _{C}, \quad \forall x,y\in C \bigl([-r,T], X_{\alpha}\bigr). $$

Let \(\phi(r)=ML_{1}r\). Then \(\phi: {\mathbb{R}}^{+}\rightarrow{\mathbb {R}}^{+}\) is continuous and nondecreasing and \(\phi(r)< r\) for \(r>0\). Hence \(Q_{1}\) is a nonlinear contraction on \(C([-r,T], X_{\alpha})\). The remaining proof is similar to the proof of Theorem 1, we omit it here. This completes the proof. □

### Remark 4

In some existing literature, see [2, 14, 16, 17], the authors always assume that \(f(t,x)\leq m(t)\) with some functions \(m\in L^{1}(J, {\mathbb{R}}^{+})\) independent of *x*, and *g* is either completely continuous or Lipschitz continuous and the coefficients satisfy some inequality conditions. But in Theorems 1 and 2, we only assume that the conditions \((H_{1})\) and \((H_{2})\) (or \((H_{2})'\)) hold. Hence, Theorems 1 and 2 greatly extend the main results of [2, 13, 14, 16, 17].