In this part, we give some properties of Green’s function associated with system (1.1)–(1.2), and we present some definitions and lemmas which are needed throughout this paper.
Definition 2.1
(see [59])
Let E be a real Banach space over \(\mathcal{R}\). A nonempty closed set \(P\subset E\) is said to be a cone provided that
-
(i)
\(au+bv\in P\) for all \(u, v\in P \) and all \(a\geq0\), \(b\geq0\) and
-
(ii)
\(u,-u\in P\) implies \(u=0\).
Every cone \(P\subset E\) induces a semi-ordering in E given by \(u\leq v\) if and only if \(v-u\in P\).
Definition 2.2
If \(x(t)=x(1-t)\), \(t\in J\), then x is said to be symmetric in J.
In our discussion, \(\mathbf{x}=(x_{1},x_{2},\ldots,x_{n})^{T}\) is symmetric on J if and only if \(x_{i}\) is symmetric on J.
Next, we reduce system (1.1)–(1.2) to an integral system. It follows from system (1.5)–(1.6) that system (1.1)–(1.2) can be written as follows:
$$ \left \{ \textstyle\begin{array}{l} (\phi_{m}(x{''}_{i}(t))){''}=\psi_{i}(t)f_{i}(t,x_{1}(t),\ldots ,x_{n}(t)), \quad0< t< 1,\\ x_{i}(0)=x_{i}(1)=\int_{0}^{1}g_{i}(s)x_{i}(s)\,ds,\\ \phi_{m}(x_{i}{''}(0))=\phi_{m}(x_{i}{''}(1))=\int_{0}^{1}h_{i}(s)\phi _{m}(x_{i}{''}(s))\,ds, \end{array}\displaystyle \right . $$
(2.1)
where \(\phi_{m}(s)=|s|^{m-2}s\), \(m>1\), \(\phi_{m^{*}}=\phi_{m}^{-1}\), \(\frac {1}{m}+\frac{1}{m^{*}}=1\).
Firstly, by means of the transformation
$$ \phi_{m}\bigl(x_{i}{''}(t) \bigr)=-y_{i}(t), $$
(2.2)
we can convert system (2.1) into
$$ \left \{ \textstyle\begin{array}{l} y_{i}{''}(t)=-\psi_{i}(t)f_{i}(t,\mathbf{x}(t)), \quad0< t< 1,\\ y_{i}(0)=y_{i}(1)=\int_{0}^{1}h_{i}(s)y_{i}(s)\,ds, \end{array}\displaystyle \right . $$
(2.3)
and
$$ \left \{ \textstyle\begin{array}{l} x_{i}{''}(t)=-\phi_{m^{*}}(y_{i}(t)), \quad 0< t< 1,\\ x_{i}(0)=x_{i}(1)=\int_{0}^{1}g_{i}(s)x_{i}(s)\,ds. \end{array}\displaystyle \right . $$
(2.4)
Lemma 2.1
Assume that
\((H_{3})\)
holds. Then system (2.4) has a unique solution
\(x_{i}(t)\)
and
\(x_{i}(t)\)
can be expressed in the form
$$ x_{i}(t)=- \int_{0}^{1}H^{i}(t,s)\phi_{m^{*}} \bigl(y_{i}(s)\bigr)\,ds, $$
(2.5)
where
$$\begin{aligned}& H^{i}(t,s)=G(t,s)+\frac{1}{1-\mu_{i}} \int_{0}^{1}G(\tau,s)g_{i}(\tau)\, d\tau, \end{aligned}$$
(2.6)
$$\begin{aligned}& G(t,s)=\left \{ \textstyle\begin{array}{l} t(1-s), \quad0\le t\le s\le1,\\ s(1-t), \quad0\le s\le t\le1. \end{array}\displaystyle \right . \end{aligned}$$
(2.7)
Proof
The proof is similar to that of Lemma 2.1 in [60]. □
By (2.6) and (2.7), we can show that \(H^{i}(t,s)\) and \(G(t,s)\) have the following properties.
Proposition 2.1
Assume that
\((H_{3})\)
holds. Then we have
$$ \begin{gathered} H^{i}(t,s)>0, \qquad G(t,s)>0,\quad \forall t,s\in(0,1); \\ H^{i}(t,s)\ge0, \qquad G(t,s)\ge0, \quad\forall t,s\in J. \end{gathered} $$
(2.8)
Proposition 2.2
For all
\(t,s\in J\), we have
$$\begin{aligned}& e(t)e(s)\le G(t,s)\le G(t,t)=t(1-t)=e(t)\le\overline{e}=\max _{t\in J}e(t)=\frac{1}{4}, \end{aligned}$$
(2.9)
$$\begin{aligned}& G(1-t,1-s)=G(t,s). \end{aligned}$$
(2.10)
Proposition 2.3
Assume that
\((H_{3})\)
holds. Then, for all
\(t,s\in J\), we have
$$ \rho^{i}e(s)\le H^{i}(t,s)\le\gamma^{i}s(1-s)= \gamma^{i}e(s)\le\frac {1}{4}\gamma^{i}, $$
(2.11)
where
$$ \gamma^{i}=\frac{1}{1-\mu_{i}},\qquad \rho^{i}=\frac{\int_{0}^{1}e(\tau )g_{i}(\tau)\,d\tau}{1-\mu_{i}}. $$
(2.12)
Proof
By (2.6) and (2.9), we have
$$ \begin{aligned}[b] H^{i}(t,s)&=G(t,s)+\frac{1}{1-\mu_{i}} \int_{0}^{1}G(s,\tau)g_{i}(\tau )\,d\tau \\ &\ge\frac{1}{1-\mu_{i}} \int_{0}^{1}G(s,\tau)g_{i}(\tau)\,d\tau \\ &\ge\frac{\int_{0}^{1}e(\tau)g_{i}(\tau)\,d\tau}{1-\mu_{i}}s(1-s) \\ &=\rho^{i}e(s), \quad t\in J. \end{aligned} $$
(2.13)
In addition, noticing that \(G(t,s)\le s(1-s)\), we have
$$ \begin{aligned}[b] H^{i}(t,s)&=G(t,s)+\frac{1}{1-\mu_{i}} \int_{0}^{1}G(s,\tau)g_{i}(\tau )\,d\tau \\ &\le s(1-s)+\frac{1}{1-\mu_{i}} \int_{0}^{1}s(1-s)g_{i}(\tau)\,d\tau \\ &\le s(1-s)\biggl[1+\frac{1}{1-\mu_{i}} \int_{0}^{1}g_{i}(\tau)\,d\tau\biggr] \\ &=s(1-s)\frac{1}{1-\mu_{i}} \\ &=\gamma^{i}e(s), \quad t\in J. \end{aligned} $$
(2.14)
□
Proposition 2.4
Assume that (\(H_{3}\)) holds. Then, for all
\(t,s\in J\), we have
$$ H^{i}(1-t,1-s)=H^{i}(t,s). $$
(2.15)
Proof
The proof is similar to that of Proposition 2.1 of [60]. □
Lemma 2.2
Assume that
\((H_{1})\)–\((H_{3})\)
hold. Then system (2.3) has a unique solution
$$ y_{i}(t)=- \int_{0}^{1}H_{1}^{i}(t,s) \psi_{i}(s)f_{i}\bigl(s,\mathbf{x}(s)\bigr)\,ds, $$
(2.16)
where
$$ H_{1}^{i}(t,s)=G(t,s)+\frac{1}{1-\nu_{i}} \int_{0}^{1}G(v,s)h_{i}(v)dv. $$
(2.17)
Proof
The proof is similar to Lemma 2.1 of [60]. □
Remark 2.1
Assume that (\(H_{3}\)) holds. Then, for all \(t,s\in J\), it follows from (2.17) that
$$H_{1}^{i}(t,s)\geq0,\qquad \rho_{1}^{i}e(s) \le H_{1}^{i}(t,s)\le\gamma _{1}^{i}s(1-s) \le\frac{1}{4}\gamma_{1}^{i},\qquad H_{1}^{i}(1-t,1-s)=H_{1}^{i}(t,s), $$
where
$$\rho_{1}^{i}=\frac{\int_{0}^{1}e(\tau)h_{i}(\tau)\,d\tau}{1-\nu_{i}}, \qquad\gamma_{1}^{i}= \frac{1}{1-\nu_{i}}. $$
Assume that \(x_{i}\) is a solution of system (2.1). Then it follows from Lemma 2.1 that
$$ x_{i}(t)=- \int_{0}^{1}H^{i}(t,s)\phi_{m^{*}} \bigl(y_{i}(s)\bigr)\,ds, $$
(2.18)
and then, it follows from Lemma 2.2 that
$$ x_{i}(t)= \int_{0}^{1}H^{i}(t,s)\phi_{m^{*}} \biggl( \int _{0}^{1}H_{1}^{i}(s,\tau) \psi_{i}(\tau)f_{i}\bigl(\tau,\mathbf{x}(\tau)\bigr)\, d\tau \biggr)\,ds. $$
(2.19)
Let \(E=C[0,1]\), \(X=\underbrace{E\times E\times\cdots\times E}_{n}\), and for all \(\mathbf{x}=(x_{1},x_{2},\ldots,x_{n})^{T}\in X\), the norm in X is defined as
$$\|\mathbf{x}\|=\sum_{i=1}^{n}\sup _{t\in J}|x_{i}|. $$
Then \((X,\|\cdot\|)\) is a real Banach space.
Define a cone K in X by
$$ \begin{aligned}[b] K={}& \Biggl\{ \mathbf{x}=(x_{1},x_{2}, \ldots,x_{n})^{T}\in X:x_{i}\ge0, x_{i}(t)\text{ is symmetric and concave on }J, \\ & \min_{t\in J}\sum_{i=1}^{n}x_{i}(t) \ge\delta\|\bf{x}\| \Biggr\} , \end{aligned} $$
(2.20)
where
$$\delta=\min_{1\le i\le n}\delta_{i},\quad \delta_{i}= \frac{\rho^{i}(\rho _{1}^{i})^{m^{*}-1}}{\gamma^{i}(\gamma_{1}^{i})^{m^{*}-1}}. $$
We also define two sets \(K_{r}\), \(K_{r,R}\) by
$$K_{r}= \{x\in K:\|x\|< r \},\qquad K_{r,R}=\{x\in K:r< \|x\|< R\}, $$
where \(0< r< R\).
To make our research significant, let \(g_{i}(t)\not\equiv0\), \(h_{i}(t)\not\equiv0\) for any \(t\in J\), \(i=1,2,\ldots,n\).
Remark 2.2
By the definition of \(\rho^{i}\), \(\rho _{1}^{i}\), \(\gamma^{i}\), \(\gamma_{1}^{i}\), we have \(0<\delta_{i}<1\), and then \(0<\delta<1\).
Let \(\mathbf{T}:K\rightarrow X\) be a map with components \((T_{1},\ldots ,T_{i}, \ldots,T_{n})\). Here, we understand \(\mathbf{Tx}= (T_{1}\mathbf{x},\ldots,T_{i}\mathbf{x},\ldots,T_{n}\mathbf{x} )^{T}\), where
$$ (T_{i}\mathbf{x}) (t)= \int_{0}^{1}H^{i}(t,s)\phi_{m^{*}} \biggl( \int _{0}^{1}H_{1}^{i}(s,\tau) \psi_{i}(\tau)f_{i}\bigl(\tau,\mathbf{x}(\tau)\bigr)\, d\tau \biggr)\,ds. $$
(2.21)
From the proof of Lemma 2.1 and Lemma 2.2, we have the following remark.
Remark 2.3
From (2.21), we know that \(\mathbf {x}\in X\) is a solution of system (1.1)–(1.2) if and only if x is a fixed point of the map \(\bf{T}\).
Lemma 2.3
Assume that
\((H_{1})\)–\((H_{3})\)
hold. Then we have
\(\mathbf{T}(K)\subset K\), and
\(\mathbf{T}:K\rightarrow K\)
is completely continuous.
Proof
For all \(\mathbf{x}\in K\), from (2.21), we know that
$$ (T_{i}\mathbf{x}){''}=-\phi_{m^{*}} \biggl( \int_{0}^{1}H_{1}^{i}(t,s)\psi _{i}(s)f_{i}\bigl(s,\mathbf{x}(s)\bigr)\,ds \biggr)\le0, $$
(2.22)
which implies that \(T_{i}\mathbf{x}\) is concave on J.
In addition, it follows from (2.21) that
$$(T_{i}\mathbf{x}) (0)=(T_{i}\mathbf{x}) (1)\ge0. $$
Thus, for all \(t\in J\), we have \((T_{i}\mathbf{x})(t)\ge0\). Noticing that \(\psi_{i}(t)\) is symmetric on \((0,1)\), \(x_{i}(t)\) is symmetric on J, and \(f_{i}(\cdot,\bf{x})\) is symmetric on J, we have
$$\begin{aligned} (T_{i}\mathbf{x}) (1-t)&= \int_{0}^{1}H^{i}(1-t,s) \phi_{m^{*}} \biggl( \int _{0}^{1}H_{1}^{i}(s,\tau) \psi_{i}(\tau)f_{i}\bigl(\tau,\mathbf{x}(\tau)\bigr)\, d\tau \biggr)\,ds \\ &= \int_{1}^{0}H^{i}(1-t,1-s) \phi_{m^{*}} \biggl( \int _{0}^{1}H_{1}^{i}(1-s, \tau)\psi_{i}(\tau)f_{i}\bigl(\tau,\mathbf{x}(\tau)\bigr)\, d\tau \biggr)\,d(1-s) \\ &= \int_{0}^{1}H^{i}(t,s)\phi_{m^{*}} \biggl( \int _{0}^{1}H_{1}^{i}(1-s, \tau)\psi_{i}(\tau)f_{i}\bigl(\tau,\mathbf{x}(\tau)\bigr)\, d\tau \biggr)\,ds \\ &= \int_{0}^{1}H^{i}(t,s)\phi_{m^{*}}\\ &\quad\times \biggl( \int _{1}^{0}H_{1}^{i}(1-s,1- \tau)\psi_{i}(1-\tau)f_{i}\bigl(1-\tau,\mathbf {x}(1-\tau) \bigr)\,d(1-\tau) \biggr)\,ds \\ &= \int_{0}^{1}H^{i}(t,s)\phi_{m^{*}} \biggl( \int_{0}^{1}H_{1}^{i}(s,\tau ) \psi_{i}(\tau)f_{i}\bigl(\tau,\mathbf{x}(\tau)\bigr)\,d\tau \biggr)\,ds \\ &=(T_{i}\mathbf{x}) (t), \end{aligned}$$
which shows that \((T_{i}\mathbf{x})(1-t)=(T_{i}\mathbf{x})(t)\), \(t\in J\). And hence \((T_{i}\mathbf{x})(t)\) is symmetric on J.
In addition, according to (2.14), we know that
$$\begin{aligned} (T_{i}\mathbf{x}) (t)&= \int_{0}^{1}H^{i}(t,s)\phi_{m^{*}} \biggl( \int _{0}^{1}H_{1}^{i}(s,\tau) \psi_{i}(\tau)f_{i}\bigl(\tau,\mathbf{x}(\tau)\bigr)\, d\tau \biggr)\,ds \\ & \le\gamma^{i}\bigl(\gamma_{1}^{i} \bigr)^{m^{*}-1} \int_{0}^{1}e(s)\phi _{m^{*}} \biggl( \int_{0}^{1}e(\tau)\psi_{i}( \tau)f_{i}\bigl(\tau,\mathbf {x}(\tau)\bigr)\,d\tau \biggr)\,ds,\quad \forall t\in J. \end{aligned} $$
Then
$$\|\mathbf{Tx}\|=\sum_{i=1}^{n}\sup _{t\in J}(T_{i}\mathbf{x}) (t)\le \sum _{i=1}^{n}\gamma^{i}\bigl( \gamma_{1}^{i}\bigr)^{m^{*}-1} \int_{0}^{1}e(s)\phi _{m^{*}} \biggl( \int_{0}^{1}e(\tau)\psi_{i}( \tau)f_{i}\bigl(\tau,\mathbf {x}(\tau)\bigr)\,d\tau \biggr)\,ds. $$
Similarly, according to (2.13), we know that
$$\begin{aligned} \min_{t\in J}\sum _{i=1}^{n}(T_{i}\mathbf{x}) (t)&=\min _{t\in J}\sum_{i=1}^{n} \int_{0}^{1}H^{i}(t,s)\phi_{m^{*}} \biggl( \int _{0}^{1}H_{1}^{i}(s,\tau) \psi_{i}(\tau)f_{i}\bigl(\tau,\mathbf{x}(\tau)\bigr)\, d\tau \biggr)\,ds \\ &\ge\sum_{i=1}^{n}\rho^{i}\bigl( \rho_{1}^{i}\bigr)^{m^{*}-1} \int_{0}^{1}e(s)\phi _{m^{*}} \biggl( \int_{0}^{1}e(\tau)\psi_{i}( \tau)f_{i}\bigl(\tau,\mathbf {x}(\tau)\bigr)\,d\tau \biggr)\,ds \\ &=\sum_{i=1}^{n}\delta_{i} \gamma^{i}\bigl(\gamma_{1}^{i} \bigr)^{m^{*}-1} \int _{0}^{1}e(s)\phi_{m^{*}} \biggl( \int_{0}^{1}e(\tau)\psi_{i}(\tau )f_{i}\bigl(\tau,\mathbf{x}(\tau)\bigr)\,d\tau \biggr)\,ds \\ &\ge\delta\|\mathbf{Tx}\|. \end{aligned} $$
Thus, we have \(T_{i}\mathbf{x}\in K\), then there is \(\mathbf {T}(K)\subset K\).
Next, we show T is completely continuous, and we need to show \(T_{i}\) is completely continuous.
Let \(l>0\) and define
$$\widehat{f}_{l}^{i}=\sup_{t\in J}\bigl\{ f_{i}\bigl(t,\mathbf{x}(t)\bigr):\mathbf{x}\in \mathcal{R}_{+}^{n}, \|\mathbf{x}\|\le l\bigr\} >0. $$
We show that \(T_{i}\) is compact.
For each \(l>0\), let \(B_{l}=\{\mathbf{x}\in K:\|\mathbf{x}\|\leq l\}\). Then \(B_{l}\) is a bounded closed convex set in K. \(\forall(\mathbf{x}_{m})_{m\in\mathcal{N}}\in K\), it follows from (2.21) that
$$\begin{aligned} |T_{i}\mathbf{x}_{m}|&= \biggl\vert \int_{0}^{1}H^{i}(t,s)\phi_{m^{*}} \biggl( \int_{0}^{1}H_{1}^{i}(s,\tau) \psi_{i}(\tau)f_{i}\bigl(\tau,\mathbf {x}_{m}( \tau)\bigr)\,d\tau \biggr)\,ds \biggr\vert \\ &\le\frac{1}{4}\gamma^{i}\biggl(\frac{1}{4} \gamma_{1}^{i}\biggr)^{m^{*}-1} \biggl\vert \int_{0}^{1}\phi_{m^{*}} \biggl( \int_{0}^{1}\psi_{i}( \tau)f_{i}\bigl(\tau ,\mathbf{x}_{m}(\tau)\bigr)\,d\tau \biggr)\,ds \biggr\vert \\ &\le\frac{1}{4}\gamma^{i}\biggl(\frac{1}{4} \gamma_{1}^{i}\biggr)^{m^{*}-1}\bigl(\widehat {f}_{l}^{i}\bigr)^{m^{*}-1} \biggl\vert \int_{0}^{1}\phi_{m^{*}} \biggl( \int _{0}^{1}\psi_{i}(\tau)\,d\tau \biggr)\,ds \biggr\vert \\ &\le\frac{1}{4}\gamma^{i}\biggl(\frac{1}{4} \gamma_{1}^{i}\biggr)^{m^{*}-1}\bigl(\widehat {f}_{l}^{i}\bigr)^{m^{*}-1} \biggl\vert \int_{0}^{1}\phi_{m^{*}}\bigl(\| \psi_{i}\| _{1} \bigr)\,ds \biggr\vert \\ &=\frac{1}{4}\gamma^{i}\biggl(\frac{1}{4} \gamma_{1}^{i}\biggr)^{m^{*}-1}\bigl(\widehat {f}_{l}^{i}\bigr)^{m^{*}-1}\bigl(\|\psi_{i} \|_{1}\bigr)^{m^{*}-1} \\ &=\biggl(\frac{1}{4}\gamma_{1}^{i} \biggr)^{m^{*}}\bigl(\widehat{f}_{l}^{i} \bigr)^{m^{*}-1}\bigl(\| \psi_{i}\|_{1}\bigr)^{m^{*}-1}. \end{aligned}$$
Therefore, \((T_{i}(B_{l}))\) is uniformly bounded.
Next we show the equicontinuity of \((T_{i}\mathbf{x}_{m})_{m\in \mathcal{N}}\). Due to \(H^{i}(t,s)\) is continuous on \(J\times J\), then \(H^{i}(t,s)\) is uniformly continuous. Thus, for any \(\varepsilon>0\), there exist \(l_{1}>0\), \(t_{1},t_{2}\in J\), if \(|t_{1}-t_{2}|< l_{1}\), we have
$$\begin{aligned} \big|(T_{i}\mathbf{x}_{m}) (t_{2})-(T_{i}\mathbf{x}_{m}) (t_{1})\big|={}& \biggl\vert \int_{0}^{1}H^{i}(t_{2},s) \phi_{m^{*}} \biggl( \int _{0}^{1}H_{1}^{i}(s,\tau) \psi_{i}(\tau)f_{i}\bigl(\tau,\mathbf{x}_{m}(\tau )\bigr)\,d\tau \biggr)\,ds \\ &- \int_{0}^{1}H^{i}(t_{1},s) \phi_{m^{*}} \biggl( \int _{0}^{1}H_{1}^{i}(s,\tau) \psi_{i}(\tau)f_{i}\bigl(\tau,\mathbf{x}_{m}(\tau )\bigr)\,d\tau \biggr)\,ds \biggr\vert \\ ={}& \biggl\vert \int_{0}^{1} \bigl[H^{i}(t_{2},s)-H^{i}(t_{1},s) \bigr]\\ &\times\phi _{m^{*}} \biggl( \int_{0}^{1}H_{1}^{i}(s,\tau) \psi_{i}(\tau)f_{i}\bigl(\tau ,\mathbf{x}_{m}( \tau)\bigr)\,d\tau \biggr)\,ds \biggr\vert \\ \le{}&\biggl(\frac{1}{4}\gamma_{1}^{i} \biggr)^{m^{*}-1}\bigl(\|\psi_{i}\| _{1}\bigr)^{m^{*}-1} \bigl(\widehat{f}_{l}^{i}\bigr)^{m^{*}-1} \int_{0}^{1} \big|H^{i}(t_{1},s)-H^{i}(t_{2},s) \big|\,ds \\ \le{}&\varepsilon, \end{aligned} $$
which shows that \((T_{i}\mathbf{x}_{m})_{m\in\mathcal{N}}\) is equicontinuous on J. Therefore, it follows from the Arzelà–Ascoli theorem that there exist a function \(T^{1}_{i}\in C[0,1]\) and a subsequence of \((T_{i}\mathbf{x}_{m})_{m\in\mathcal{N}}\) converging uniformly to \(T^{1}_{i}\) on J.
We prove the continuity of \(T_{i}\). Let \((\mathbf{x}_{m})_{m\in \mathcal{N}}\) be any sequence converging on K to \(\mathbf{x}\in K\), and let \(L>0\) be such that \(\|\mathbf{x}_{m}\|\le L\) for all \(m\in \mathcal{N}\). Note that \(f_{i}(t,\mathbf{x})\) is continuous on \(J\times K_{L}\). It is not difficult to see that the dominated convergence theorem guarantees that
$$ \lim_{m\rightarrow\infty}(T_{i}\mathbf{x}_{m}) (t)=(T_{i}\mathbf {x}) (t) $$
(2.23)
for each \(t\in J\). Moreover, the compactness of \(T_{i}\) implies that \((T_{i}\mathbf{x}_{m})(t)\) converges uniformly to \((T_{i}\mathbf {x})(t)\) on J. If not, then there exist \(\varepsilon_{0}>0\) and a subsequence \((\mathbf{x}_{m_{j}})_{j\in\mathcal{N}}\) of \((\mathbf {x}_{m})_{m\in\mathcal{N}}\) such that
$$ \sup_{t\in J}\big|(T_{i}\mathbf{x}_{m_{j}}) (t)-(T_{i}\mathbf{x}) (t)\big|\ge \varepsilon_{0}, \quad j\in \mathcal{N}. $$
(2.24)
Now, it follows from the compactness of \(T_{i}\) that there exists a subsequence of \((\mathbf{x}_{m_{j}})_{j\in\mathcal{N}}\) (without loss of generality, assume that the subsequence is \((\mathbf {x}_{m_{j}})_{j\in\mathcal{N}}\)) such that \((T_{i}\mathbf {x}_{m_{j}})_{j\in\mathcal{N}}\) converges uniformly to \(y_{0}\in C[0,1]\). Thus, from (2.24), we easily see that
$$ \sup_{t\in J}\big|y_{0}(t)-(T_{i}\mathbf{x}) (t)\big|\ge\varepsilon_{0},\quad j\in \mathcal{N}. $$
(2.25)
On the other hand, from the pointwise convergence (2.23) we obtain
$$y_{0}(t)=(T_{i}\mathbf{x}) (t),\quad t\in J. $$
This is a contradiction to (2.25). Therefore \(T_{i}\) is continuous.
Therefore \(T_{i}:K\rightarrow K\) is completely continuous. This completes the proof of Lemma 2.3. □
In the following lemma, we employ Hölder’s inequality to obtain some of the norm inequalities in our main results.
Lemma 2.4
(Hölder)
Let
\(e\in L^{p}[a,b]\)
with
\(p>1\), \(h\in L^{q}[a,b]\)
with
\(q>1\), and
\(\frac{1}{p}+\frac{1}{q}=1\). Then
\(eh\in L^{1}[a,b]\), and
$$\|eh\|_{1}\le\|e\|_{p}\|h\|_{q}. $$
Let
\(e\in L^{1}[a,b]\)
and
\(h\in L^{\infty}[a,b]\). Then
\(eh\in L^{1}[a,b]\), and
$$\|eh\|_{1}\le\|e\|_{1}\|h\|_{\infty}. $$
Finally, we state the well-known fixed point theorem of cone expansion and compression of norm type.
Lemma 2.5
(see [59])
Let
P
be a cone in a real Banach space
E. Assume
\(\Omega_{1}\), \(\Omega_{2}\)
are bounded open sets in
E
with
\(0\in\overline{\Omega}_{1}\subset\Omega_{2}\), \(A:P\cap (\overline{\Omega}_{2}\setminus\Omega_{1})\rightarrow P\)
is completely continuous such that either
-
(i)
\(\|Ax\|\le\|x\|\), \(\forall x\in P\cap\partial\Omega_{1}\); \(\|Ax\| \ge\|x\|\), \(\forall x\in P\cap\partial\Omega_{2}\)
or
-
(ii)
\(\|Ax\|\ge\|x\|\), \(\forall x\in P\cap\partial\Omega_{1}\); \(\|Ax\| \le\|x\|\), \(\forall x\in P\cap\partial\Omega_{2}\).
Then
A
has at least one fixed point in
\(P\cap(\overline{\Omega }_{2}\setminus\Omega_{1})\).
Remark 2.4
To make it clear for the reader what \(\Omega _{1}\), \(\Omega_{2}\), \(\partial\Omega_{1}\), \(\partial\Omega_{2}\), and \(\bar{\Omega}_{2}\setminus\Omega_{1}\) mean, we give typical examples of \(\Omega_{1}\) and \(\Omega_{2}\).
$$\begin{gathered} \Omega_{1}=\bigl\{ x\in C[a,b]:\|x\|< r\bigr\} ,\qquad \Omega_{2}= \bigl\{ x\in C[a,b]:\|x\|< R\bigr\} , \\ \overline{\Omega}_{2}\setminus\Omega_{1}=\bigl\{ x\in C[a,b]:r\leq\|x\|\leq R\bigr\} ,\end{gathered} $$
where \(0< r< R\), \(\|x\|=\max_{t\in[a,b]}|x(t)|\).