Fixed point theorems of convex-power 1-set-contraction operators in Banach spaces

Abstract

In this article, we give the definition of a class of new operators, namely, convex-power 1-set-contraction operators in Banach spaces, and study the existence of fixed points of this class of operators. By using methods of approximation by operators, we obtain fixed point theorems of convex-power 1-set-contraction operators, which generalize fixed point theorems of 1-set-contraction operators in Banach spaces. By using the fixed point theorem, the existence of solutions of nonlinear Sturm-Liouville problems in Banach spaces is investigated under more general conditions than those used in former literatures.

Mathematics Subject Classification 2010: 47H10.

0 Introduction

For the need of studying differential equations and integral equations, Sun and Zhang [1] gave the definition of convex-power condensing operators and obtained the fixed point theorem of this class of operators. Li [2] gave the fixed point theorem of semi-closed 1-set-contraction operators.

In this article, by combinating the definitions of convex-power condensing operators and 1-set-contraction operators, we give the definition of convex-power 1-set-contraction operators in Banach spaces and study the existence of fixed points of this class of new operators. The results in this article generalize the ones in [1–3]. By using the fixed point theorem, the existence of solutions of nonlinear Sturm-Liouville problems in Banach spaces is investigated under more general conditions than those used in former literatures.

1 Preliminaries

Before providing the main results, we introduce some basic definitions and results (see [1–6]).

In this article, we always assume that E is a Banach space, D ⊂ E, and α(S) denotes the Kuratowski measure of noncompactness of a bounded set S ⊂ E.

Let A: D → E be continuous. If there exists a constant k ≥ 0, such that for any bounded subset S ⊂ D,

$$\alpha \left(A\left(S\right)\right)\le k\alpha \left(S\right).$$

Then A is said to be k-set-contraction in D.

Let A: D → E be continuous and bounded. If there exist x₀ ∈ D and a positive integer n₀, such that for any bounded and nonprecompact subset S ⊂ D,

$$\alpha \left(A^{\left(n_0,x_0\right)}\left(S\right)\right)<\alpha \left(S\right),$$

where $A^{\left(1,x_0\right)}\left(S\right)=A\left(S\right)$, $A^{\left(n,x_0\right)}\left(S\right)=A\left(\stackrel{-}{co}\left\{A^{\left(n-1,x_0\right)}\left(S\right),x_0\right\}\right)$, n = 2, 3,....

Then A is said to be convex-power condensing.

Lemma 1.1[1]. Let D ⊂ E be bounded, convex, and closed. Suppose that A: D → D is convex-power condensing, then A has at least one fixed point in D.

Definition 1.1[2]. A: D → E is said to be semi-closed if for any closed set F ⊂ D, (I - A)F is closed.

Definition 1.2[3]. Let A: D → E, {x _n } ⊂ D bounded, {x _n -Ax _n } strongly convergent. A is said to be semi-compact if {x _n } has a strongly convergent subsequence.

2 Main results

Next, we will give the definition of convex-power 1-set-contraction operators in Banach spaces.

Definition 2.1. Let A : D → E be continuous and bounded. If there exist x₀ ∈ D and a positive integer n₀, such that for any bounded subset S ⊂ D,

$$\alpha \left(A^{\left(n_0,x_0\right)}\left(S\right)\right)\le \alpha \left(S\right),$$

where$A^{\left(1,x_0\right)}\left(S\right)=A\left(S\right)$, $A^{\left(n,x_0\right)}\left(S\right)=A\left(\stackrel{-}{co}\left\{A^{\left(n-1,x_0\right)}\left(S\right),x_0\right\}\right)$, n = 2, 3,....

Then A is said to be convex-power 1-set-contraction in Banach spaces.

Remark 2.1. Obviously, 1-set-contraction operators are convex-power 1-set-contraction operators. Convex-power 1-set-contraction operators are more general.

Now, we establish the main theorem as follows:

Theorem 2.1. Let E be a Banach space, D ⊂ E bounded, convex, and closed. Suppose that A: D → D is semi-closed and convex-power 1-set-contraction, then A has at least one fixed point in D.

Proof. Since A is convex-power 1-set-contraction, there exist x₀ ∈ D and a positive integer n₀, such that for any bounded subset S ⊂ D,

$$\alpha \left(A^{\left(n_0,x_0\right)}\left(S\right)\right)\le \alpha \left(S\right).$$

$\forall x\in D,\text{let}{A}_{n}x=\left(1-\frac{1}{n}\right)Ax+\frac{1}{n}{x}_0\left(n=2,3,\dots \right)$, then A _n : D → D. ∀y ∈ D - x₀ = {x - x₀|x ∈ D}, let By = A(y + x₀)-x₀.

For any bounded subset S ⊂ D,

$$\begin{array}{c}{B}^{\left(1,0\right)}\left(S-x_0\right)=B\left(S-x_0\right)=A\left(S\right)-x_0=A^{\left(1,x_0\right)}\left(S\right)-x_0,\\ A_n^{\left(1,x_0\right)}\left(S\right)=A_n\left(S\right)=\left(1-\frac{1}{n}\right)A\left(S\right)+\frac{1}{n}{x}_0=\left(1-\frac{1}{n}\right)A^{\left(1,x_0\right)}\left(S\right)+\frac{1}{n}{x}_0;\\ B^{\left(2,0\right)}\left(S-x_0\right)=B\left(\stackrel{-}{co}\left\{B^{\left(1,0\right)}\left(S-x_0\right),0\right\}\right)=B\left(\stackrel{-}{co}\left\{A^{\left(1.x_0\right)}\left(S\right)-x_0,0\right\}\right)\\ =A\left(\stackrel{-}{co}\left\{A^{\left(1,x_0\right)}\left(S\right)-x_0,0\right\}+x_0\right)-x_0\hfill \\ =A\left(\stackrel{-}{co}\left\{A^{\left(1,x_0\right)}\left(S\right),x_0\right\}\right)-x_0=A^{\left(2,x_0\right)}\left(S\right)-x_0,\end{array}$$

$$\begin{array}{cc}\hfill A_n^{\left(2,x_0\right)}\left(S\right)& =A_n\left(\stackrel{-}{co}\left\{A_n^{\left(1,x_0\right)}\left(S\right),x_0\right\}\right)\hfill \\ =A_n\left(\stackrel{-}{co}\left\{\left(1-\frac{1}{n}\right)B^{\left(1,0\right)}\left(S-x_0\right)+x_0,x_0\right\}\right)\hfill \\ =A_n\left(\stackrel{-}{co}\left\{\left(1-\frac{1}{n}\right)B^{\left(1,0\right)}\left(S-x_0\right),0\right\}+x_0\right)\hfill \\ \subset A_n\left(\stackrel{-}{co}\left\{B^{\left(1,0\right)}\left(S-x_0\right),0\right\}+x_0\right)\hfill \\ =A_n\left(\stackrel{-}{co}\left\{B^{\left(1,0\right)}\left(S-x_0\right)+x_0,x_0\right\}\right)\hfill \\ =A_n\left(\stackrel{-}{co}\left\{A^{\left(1,x_0\right)}\left(S\right),x_0\right\}\right)=\left(1-\frac{1}{n}\right)A^{\left(2,x_0\right)}\left(S\right)+\frac{1}{n}{x}_0;\hfill \end{array}$$

and generally,

$$\begin{array}{cc}\hfill B^{\left(n_0,0\right)}\left(S-x_0\right)& =B\left(\stackrel{-}{co}\left\{B^{\left(n_0-1,0\right)}\left(S-x_0\right),0\right\}\right)=B\left(\stackrel{-}{co}\left\{A^{\left(n_0-1,x_0\right)}\left(S\right)-x_0,0\right\}\right)\hfill \\ =A\left(\stackrel{-}{co}\left\{A^{\left(n_0-1,x_0\right)}\left(S\right)-x_0,0\right\}+x_0\right)-x_0=A^{\left(n_0,x_0\right)}\left(S\right)-x_0,\hfill \end{array}$$

$$\begin{array}{cc}\hfill A_n^{\left(n_0,x_0\right)}\left(S\right)& =A_n\left(\stackrel{-}{co}\left\{A_n^{\left(n_0-1,x_0\right)}\left(S\right),x_0\right\}\right)\hfill \\ \subset A_n\left(\stackrel{-}{co}\left\{\left(1-\frac{1}{n}\right)A^{\left(n_0-1,x_0\right)}\left(S\right)+\frac{1}{n}{x}_0,x_0\right\}\right)\hfill \\ =A_n\left(\stackrel{-}{co}\left\{\left(1-\frac{1}{n}\right)B^{\left(n_0-1,0\right)}\left(S-x_0\right)+x_0,x_0\right\}\right)\hfill \\ \subset A_n\left(\stackrel{-}{co}\left\{B^{\left(n_0-1,0\right)}\left(S-x_0\right),0\right\}+x_0\right)\hfill \\ =A_n\left(\stackrel{-}{co}\left\{A^{\left(n_0-1,x_0\right)}\left(S\right),x_0\right\}\right)\hfill \\ =\left(1-\frac{1}{n}\right)A^{\left(n_0,x_0\right)}\left(S\right)+\frac{1}{n}{x}_0.\hfill \end{array}$$

By the definition of the convex-power 1-set-contraction operator and the properties of the measure of noncompactness, we have

$$\alpha \left(A_n^{\left(n_0,x_0\right)}\left(S\right)\right)\le \left(1-\frac{1}{n}\right)\alpha \left(A^{\left(n_0,x_0\right)}\left(S\right)\right)\le \left(1-\frac{1}{n}\right)\alpha \left(S\right)<\alpha \left(S\right),n=2,3,\dots .$$

Therefore, A _n : D → D is convex-power condensing. By Lemma 1.1, A _n has a fixed point x _n in D, i.e., A _n x _n = x _n (n = 2, 3,...). Since $\left|\right|Ax-A_{n}x\left|\right|=\frac{1}{n}\left|\right|Ax-x_0\left|\right|$, ∀x ∈ D, and A is bounded in D, then for any x ∈ D, ||Ax - A _n x|| → 0 (n → +∞). Obviously,

$$\left|\right|A{x}_n-x_n\left|\right|=\left|\right|A{x}_n-A_n{x}_n\left|\right|\to 0\left(n\to +\infty \right).$$

i.e., Ax _n - x _n → 0 (n → +∞). Since A is semi-closed and D is closed, 0 ∈ (I - A)D. Therefore, there exists x₀ ∈ D, such that x₀ = Ax₀. The proof is completed.

Remark 2.2. In Theorem 2.1, let n₀ = 1, the fixed point theorem of semiclosed 1-set-contraction operators in[2]is obtained. Therefore, Theorem2.1. generalizes the fixed point theorem of semi-closed 1-set-contraction operators.

Theorem 2.2. Let E be a Banach space, D ⊂ E bounded, convex and closed. Suppose that A: D → D is semi-compact and convex-power 1-set-contraction, then A has at least one fixed point in D.

Proof. $\forall x\in D,\text{let}{A}_{n}x=\left(1-\frac{1}{n}\right)Ax+\frac{1}{n}{x}_0\left(n=2,3,\dots \right)$. By the proof of Theorem 2.1, A _n : D → D has a fixed point x _n , and {A _n } is uniformly convergent to A in D. By x _n ∈ D, then ||A _n x _n - Ax _n || → 0. i.e., ||x _n - Ax _n || → 0. Therefore, (I - A)(x _n ) → 0 (n → +∞).

Since A is semi-compact and {x _n } ⊂ D is bounded, {x _n } has a convergent subsequence $\left\{x_{n_i}\right\}$. Let $x_{n_i}\to x_0\left(n_i\to +\infty \right)$. Since D is closed, x₀ ∈ D. Since A is continuous in D, $x_{n_i}-A{x}_{n_i}\to x_0-A{x}_0\left(n_i\to +\infty \right)$. By $x_{n_i}-A{x}_{n_i}\to 0\left(n_i\to +\infty \right)$, we have x₀ - Ax₀ = 0. The proof is completed.

3 Application

Let E be a Banach space. Consider the existence of solutions of nonlinear Sturm-Liouville problems in E as follows:

$$\left\{\begin{array}{cc}\hfill -\left(Lx\right)\left(t\right)=f\left(t,x\right),\hfill & \hfill t\in \left(0,1\right);\hfill \\ \hfill ax\left(0\right)-b{x}^{\prime }\left(0\right)=0,\hfill & \hfill cx\left(1\right)+d{x}^{\prime }\left(1\right)=0\hfill \end{array}\right.$$

(3.1)

where (Lx)(t) = (p(t)x')'+q(t)x, f ∈ C[I × E, E](I = [0, 1]).

Assume that

$$\begin{array}{cc}\hfill \left({\text{H}}_1\right)& p\left(t\right)\in C^1\left[I,R\right],p\left(t\right)>0,q\left(t\right)\in C\left[I,R\right],q\left(t\right)\le 0,\hfill \\ a\ge 0,b\ge 0,c\ge 0,d\ge 0,a^2+b^2\ne 0,c^2+d^2\ne 0,\hfill \end{array}$$

and the homogeneous equations of (3.1)

$$\left\{\begin{array}{cc}-\left(Lx\right)\left(t\right)=0,\hfill & t\in \left(0,1\right);\hfill \\ ax\left(0\right)-b{x}^{\prime }\left(0\right)=0,\hfill & cx\left(1\right)+d{x}^{\prime }\left(1\right)=0\hfill \end{array}\right.$$

(3.2)

has only zero solution in C² [I, R].

Let G(t, s) be Green function of (3.2), i.e.,

$$G\left(t,s\right)=\left\{\begin{array}{cc}\hfill \frac{1}{\rho }u\left(t\right)v\left(s\right),\hfill & \hfill 0\le t\le s\le 1;\hfill \\ \hfill \frac{1}{\rho }u\left(s\right)v\left(t\right),\hfill & \hfill 0\le s\le t\le 1.\hfill \end{array}\right.$$

(3.3)

Lemma 3.1[6]. Assume that (H₁) holds, then Green function G(t, s) of (3.3) has the following properties:

(i) G(t, s) is continuous and symmetric in [0, 1] × [0, 1];

(ii) u(t) ∈ C²[0, 1] is monotonically increasing, and u(t) > 0, t ∈ (0, 1];

(iii) v(t) ∈ C²[0, 1] is monotonically decreasing, and v(t) > 0, t ∈ [0, 1);

(iv) (Lu)(t) ≡ 0, u(0) = b, u'(0) = a;

(v) (Lv)(t) ≡ 0, v(0) = d, v'(0) = -c;

(vi) ρ is a positive constant.

Let

$$\begin{array}{c}\left(Tx\right)\left(t\right)={\int }_0^1\begin{array}{ccc}\hfill G\left(t,s\right)f\left(s,x\left(s\right)\right)ds,\hfill & \hfill t\in \left[0,1\right],\hfill & \hfill x\in C\left[I,E\right],\hfill \end{array}\\ \left(K\phi \right)\left(t\right)={\int }_0^1\begin{array}{ccc}\hfill G\left(t,s\right)\phi \left(s\right)ds,\hfill & \hfill t\in \left[0,1\right],\hfill & \hfill \phi \in C\left[I,R\right].\hfill \end{array}\end{array}$$

We can prove that the solution in C²[I, E] of (3.1) is equivalent to the fixed point of T (see [7]).

Since G(t, s) is continuous, it can be easily proved that K: C [I, R] → C [I, R] is linear and completely continuous. By Lemma 3.1, ∀t, s ∈ [0, 1],

$$\frac{u\left(s\right)v\left(s\right)}{u\left(1\right)v\left(0\right)}G\left(t,t\right)\le G\left(t,s\right)\le G\left(t,t\right).$$

Therefore, by Krein-Rutman Theorem [6], the first characteristic value of K is λ₁ > 0, and λ₁ = (r(K))^-1.

Now we give some conditions:

(H ₂ ) f ∈ C[I × E, E], for any bounded subset B in E, f is uniformly continuous in I × B, and there exists k ∈ [0, λ₁), such that

$$\begin{array}{cc}\hfill \alpha \left(f\left(t,B\left(t\right)\right)\right)\le k\alpha \left(B\left(t\right)\right),\hfill & \hfill \forall t\in \left[0,1\right],\hfill \end{array}$$

where λ₁ is the first characteristic value of K.

(H ₃ ) there exist M ∈ (0, λ₁) and h(t) ∈ C[I , R⁺], such that for any (t, x) ∈ I × E,

$$\left|\right|f\left(t,x\right)\left|\right|\le M\left|\right|x\left|\right|+h\left(t\right).$$

Theorem 3.1. Suppose that (H₁), (H₂), (H₃) hold, then Sturm-Liouville problems (3.1) has at least one solution in C²[I, E].

To prove Theorem 3.1, here we introduce some lemmas.

Lemma 3.2[7]. For M < λ₁as above, let K₁ = M K, then there exists a norm$\left|\right|\cdot |{|}_{C\left[I,R\right]}^{*}$which is equivalent to || · ||_{C[I, R]}and satisfies:

(1)$\left|\right|K_1\phi |{|}_{C\left[I,R\right]}^{*}\le \sigma |\left|\phi \right|{|}_{C\left[I,R\right]}^{*}$, where$\sigma =\frac{M+{\lambda }_1}{2{\lambda }_1}$,

(2) if 0 ≤ φ(t) ≤ ψ(t), ∀ t ∈ I, then$\left|\right|\phi |{|}_{C\left[I,R\right]}^{*}\le |\left|\psi \right|{|}_{C\left[I,R\right]}^{*}$, where ||φ||_{C[I, R]}= max_t∈I|φ(t)|.

Lemma 3.3[7]. If B ⊂ C [I, E] is equicontinuous, u₀ ∈ C[I, E], then$\stackrel{-}{co}\left\{B,u_0\right\}$is also equicontinuous in C[I, E].

Lemma 3.4[7]. If B ⊂ C[I, E] is equicontinuous and bounded, then α(B) = max_t∈Iα(B(t)).

Lemma 3.5[7]. If B ⊂ C[I, E] is equicontinuous and bounded, then α(B(t)) ∈ C[I, R⁺], and

$$\alpha \left({\int }_{t_0}^{t}B\left(s\right)ds\right)\le {\int }_{t_0}^t\begin{array}{cc}\hfill \alpha \left(B\left(s\right)\right)ds,\hfill & \hfill \forall t\in I.\hfill \end{array}$$

Proof of Theorem 3.1. Set $R_1>\frac{2{\lambda }_1}{{\lambda }_1-M}\left|\right|Kh|{|}_{C\left[I,R\right]}^{*}$, where $\left(Kh\right)\left(t\right)={\int }_0^{1}G\left(t,s\right)h\left(s\right)ds$.

Let $D=\left\{x\in C\left[I,E\right]|\phi \left(t\right)=|\left|x\left(t\right)\right|\left|\text{and}\right|\left|\phi \right|{|}_{C\left[I,R\right]}^{*}\le R_1\right\}$. Since ||·||_{C[I, R]}is equivalent to $\left|\right|\cdot |{|}_{C\left[I,R\right]}^{*}$, D is bounded, convex, and closed in C[I, E].

First ∀x ∈ D, ||x||_{C[I, E]}= max_t∈I||x(t)|| = max_t∈Iφ(t) = max_t∈I|φ(t)| = ||φ||_{C[I, R]}, then D is bounded.

Second, ∀x _n ∈ D, x _n →x₀, n → +∞. Therefore, ${\phi }_n\left(t\right)=\left|\right|x_n\left(t\right)\left|\right|,\left|\right|{\phi }_n|{|}_{C\left[I,R\right]}^{*}\le R_1,||x_n-x_0|{|}_{C\left[I,E\right]}\to 0$, i.e., max_t∈I||x _n (t) - x₀(t)|| → 0.

Let ${\tilde{\phi }}_n\left(t\right)=\left|\right|x_n\left(t\right)-x_0\left(t\right)\left|\right|$, φ₀(t) = ||x₀(t)||, then ${\phi }_0\left(t\right)\le {\tilde{\phi }}_n\left(t\right)+{\phi }_n\left(t\right),\left|\right|{\tilde{\phi }}_n|{|}_{C\left[I,R\right]}\to 0$. By Lemma 3.2,

$$\begin{array}{cc}\hfill \left|\right|{\phi }_0|{|}_{C\left[I,R\right]}^{*}& \le \left|\right|{\tilde{\phi }}_n+{\phi }_n|{|}_{C\left[I,R\right]}^{*}\hfill \\ \le \left|\right|{\tilde{\phi }}_n|{|}_{C\left[I,R\right]}^{*}+|\left|{\phi }_n\right|{|}_{C\left[I,R\right]}^{*}\hfill \\ \le \left|\right|{\tilde{\phi }}_n|{|}_{C\left[I,R\right]}^{*}+R_1\hfill \end{array}$$

Let n → +∞, then $\left|\right|{\phi }_0|{|}_{C\left[I,R\right]}^{*}\le R_1$, i.e., x₀ ∈ D, D is closed.

Finally, ∀x₁, x₂ ∈ D, 0 ≤ α ≤ 1. Let φ _i (t) = ||x _i (t)||, i = 1,2; φ₃(t) = ||αx₁(t)+(1-α)x₂(t)||. Obviously, φ₃ ≤ αφ₁(t)+(1-α)φ₂(2). By Lemma 3.2,

$$\left|\right|{\phi }_3\left|{|}_{C\left[I,R\right]}^{*}\right||\le \alpha |\left|{\phi }_1\right|{|}_{C\left[I,R\right]}^{*}\left|\right|+\left(1-\alpha \right)\left|\right|{\phi }_2|{|}_{C\left[I,R\right]}^{*}\le R_1.$$

Then D is convex. Therefore, D is bounded, convex, and closed.

By (H₂), f is uniformly continuous in I × D, then T: D → C[I, E] is continuous.

First, we prove that T: D → D. For any given x in D, let φ(t) = ||Tx(t)||ψ(t) = ||x(t)||. By (H₃),

$$\begin{array}{cc}\hfill \phi \left(t\right)& =\left|\right|Tx\left(t\right)\left|\right|=\left|\right|{\int }_0^{1}G\left(t,s\right)f\left(s,x\left(s\right)\right)ds\left|\right|\hfill \\ \le {\int }_0^{1}G\left(t,s\right)\left|\right|f\left(s,x\left(s\right)\right)\left|\right|ds\hfill \\ \le {\int }_0^{1}G\left(t,s\right)M\psi \left(s\right)ds+{\int }_0^{1}G\left(t,s\right)h\left(s\right)ds\hfill \\ =\left(K_1\psi \right)\left(t\right)+\left(Kh\right)\left(t\right).\hfill \end{array}$$

By Lemma 3.2,

$$\begin{array}{cc}\hfill \left|\right|\phi |{|}_{C\left[I,R\right]}^{*}& \le \left|\right|K_1\psi +Kh|{|}_{C\left[I,R\right]}^{*}\hfill \\ \le \left|\right|K_1\psi |{|}_{C\left[I,R\right]}^{*}+|\left|Kh\right|{|}_{C\left[I,R\right]}^{*}\hfill \\ \le \sigma \left|\right|\psi |{|}_{C\left[I,R\right]}^{*}+|\left|Kh\right|{|}_{C\left[I,R\right]}^{*}\hfill \\ \le \sigma R_1+\frac{{\lambda }_1-M}{2{\lambda }_1}{R}_1=R_1.\hfill \end{array}$$

Therefore, T : D → D is continuous

Next, we prove that T(D) is equicontinuous in C[I, E]. By (H₂), ∃M₁ > 0, ||f(t, x)|| ≤ M₁, ∀(t, x) ∈ I × D. Then,

$$\left|\right|Tx\left(t_1\right)-Tx\left(t_2\right)\left|\right|\le M_1{\int }_0^1\begin{array}{cc}\hfill |G\left(t_1,s\right)-G\left(t_2,s\right)|ds,\hfill & \hfill \forall t_1,t_2\in I,x\in D.\hfill \end{array}$$

Therefore, T(D) is equicontinuous.

Let $F=\stackrel{-}{co}T\left(D\right)\subset D$. Obviously, F is bounded, convex, and closed, and $T\left(\stackrel{-}{co}T\left(D\right)\right)\subset T\left(D\right)\subset \stackrel{-}{co}T\left(D\right)$, i.e., T: F → F. By Lemma 3.3, F is equicontinuous in C[I, E].

Next, we prove that T: F → F is convex-power 1-set-contraction. Obviously, T is bounded and continuous. Set x₀ ∈ F, we'll prove that there exists n₀, such that for any bounded B ⊂ F,

$$\alpha \left(T^{\left(n_0,x_0\right)}\left(B\right)\right)\le \alpha \left(B\right).$$

By B ⊂ F ⊂ D, T (B) is equicontinuous. Then $T^{\left(2,x_0\right)}\left(B\right)$ is equicontinuous from $T^{\left(2,x_0\right)}\left(B\right)=T\left(\stackrel{-}{co}\left\{T\left(B\right),x_0\right\}\right)\subset T\left(D\right)$ Generally, ∀ n ∈ N, $T^{\left(n,x_0\right)}\left(B\right)$ is equicontinuous. Since $T^{\left(n,x_0\right)}\left(B\right)$ is bounded, By Lemma 3.4,

$$\alpha \left(T^{\left(n,x_0\right)}\left(B\right)\right)=\underset{t\in I}{\text{max}}\alpha \left(\left(T^{\left(n,x_0\right)}\left(B\right)\right)\left(t\right)\right)n=2,3,....$$

(3.4)

Since G(t, s) is continuous in I ×I, f is uniformly continuous in I ×D, then

$$\begin{array}{c}\left|\right|G\left(t,s_1\right)f\left(s_1,x\left(s_1\right)\right)-G\left(t,s_2\right)f\left(s_2,x\left(s_2\right)\right)\left|\right|\\ \le \left|\right|G\left(t,s_1\right)-G\left(t,s_2\right)\left|\right|\left|\right|f\left(s_1,x\left(s_1\right)\right)\left|\right|+\left|\right|G\left(t,s_2\right)\left|\right|\left|\right|f\left(s_1,x\left(s_1\right)\right)-f\left(s_2,x\left(s_2\right)\right)\left|\right|\\ \left(\forall s_1,s_2\in I,x\in B\right)\end{array}$$

Therefore G(t, s)f(s, B(s))(∀ s, t ∈ I) is equicontinuous in C[I, E]. By (H₂), Lemmas 3.4 and 3.5,

$$\begin{array}{cc}\hfill \alpha \left(\left(T^{\left(1,x_0\right)}\left(B\right)\right)\left(t\right)\right)& =\alpha \left(\left(T\left(B\right)\right)\left(t\right)\right)\hfill \\ =\alpha \left({\int }_0^{1}G\left(t,s\right)f\left(s,B\left(s\right)\right)ds\right)\hfill \\ \le {\int }_0^{1}G\left(t,s\right)\alpha \left(f\left(s,B\left(s\right)\right)\right)ds\hfill \\ \le k{\int }_0^{1}G\left(t,s\right)\alpha \left(B\left(s\right)\right)ds\hfill \\ \le k\alpha \left(B\right){\int }_0^{1}G\left(t,s\right)ds\hfill \\ =k\alpha \left(B\right)\cdot K{\phi }_0\left(t\right)\hfill \end{array}$$

where φ₀(t) ≡ 1, ∀ t ∈ I.

By the equicontinuity of $T^{\left(1,x_0\right)}\left(B\right)=T\left(B\right)$ and the uniform continuity of f, $G\left(t,s\right)f\left(s,\stackrel{-}{co}\left\{\left(T^{\left(1,x_0\right)}\left(B\right)\right)\left(s\right),x_0\right\}\right)\left(\forall s,t\in I\right)$ is equicontinuous. Therefore,

$$\begin{array}{cc}\hfill \alpha \left(\left(T^{\left(2,x_0\right)}\left(B\right)\right)\left(t\right)\right)& =\alpha \left(T\stackrel{-}{co}\left\{\left(T^{\left(1,x_0\right)}\left(B\right)\right)\left(t\right),x_0\right\}\right)\hfill \\ =\alpha \left({\int }_0^{1}G\left(t,s\right)f\left(s,\stackrel{-}{co}\left\{\left(T^{\left(1,x_0\right)}\left(B\right)\right)\left(s\right),x_0\right\}\right)ds\right)\hfill \\ \le {\int }_0^{1}G\left(t,s\right)\alpha \left(f\left(s,\stackrel{-}{co}\left\{\left(T^{\left(1,x_0\right)}\left(B\right)\right)\left(s\right),x_0\right\}\right)\right)ds\hfill \\ \le k{\int }_0^{1}G\left(t,s\right)\alpha \left(\stackrel{-}{co}\left\{\left(T^{\left(1,x_0\right)}\left(B\right)\right)\left(s\right),x_0\right\}\right)ds\hfill \\ =k{\int }_0^{1}G\left(t,s\right)\alpha \left(\left(T^{\left(1,x_0\right)}\left(B\right)\right)\left(s\right)\right)ds\hfill \\ \le k^2\alpha \left(B\right){\int }_0^{1}G\left(t,s\right)K{\phi }_0\left(s\right)ds\hfill \\ =k^2\alpha \left(B\right)\cdot K^2{\phi }_0\left(t\right).\hfill \end{array}$$

Generally,

$$\alpha \left(\left(T^{\left(n,x_0\right)}\left(B\right)\right)\left(t\right)\right)\le k^n\alpha \left(B\right)\cdot K^n{\phi }_0\left(t\right)$$

We have $r\left(kK\right)=kr\left(K\right)=k\cdot {\lambda }_1^{-1}<{\lambda }_1\cdot {\lambda }_1^{-1}=1$. By the definition of spectral radius, let $\epsilon =\frac{1-r\left(kK\right)}{2}$, then ∃ m₀ > 0, when n > m₀,

$$\begin{array}{cc}\hfill {\text{max}}_{t\in I}\left|k^n{K}^n{\phi }_0\left(t\right)\right|& =\left|\right|k^n{K}^n{\phi }_0\left|\right|\hfill \\ \le \left|\right|k^n{K}^n\left|\right|\left|\right|{\phi }_0\left|\right|=\left|\right|k^n{K}^n\left|\right|\hfill \\ \le {\left(r\left(kK\right)+\epsilon \right)}^n={\left(\frac{1+r\left(kK\right)}{2}\right)}^n<1.\hfill \end{array}$$

Set n₀ > m₀, then ∀t ∈ I,

$$\begin{array}{cc}\hfill \alpha \left(\left(T^{\left(n_0,x_0\right)}\left(B\right)\right)\left(t\right)\right)& \le k^{n_0}\alpha \left(B\right)\cdot K^{n_0}{\phi }_0\left(t\right)\hfill \\ \le \left|\right|k^{n_0}\cdot K^{n_0}{\phi }_0\left|\right|\alpha \left(B\right)\hfill \\ \le {\left(\frac{1+r\left(kK\right)}{2}\right)}^{n_0}\alpha \left(B\right)\le \alpha \left(B\right).\hfill \end{array}$$

By (3.4), $\alpha \left(T^{\left(n_0,x_0\right)}\left(B\right)\right)\le \alpha \left(B\right)$. Therefore, T: F → F is convex-power 1-set-contraction. Since f is uniformly continuous, T is semi-closed. By Theorem 2.1, T has one fixed point in C[I, E], i.e., Sturm-Liouville problems (3.1) has at least one solution in C²[I, E].