Fixed point theorems for a class of nonlinear operators in Hilbert spaces with lattice structure and application

Abstract

We discuss the existence of a fixed point for a class of nonlinear operators in Hilbert spaces with lattice structure by a combination of variational and partial ordered methods. An application to second-order ordinary differential equations is included.

1 Introduction

Let E be a Banach space with a cone P. Then E becomes an ordered Banach space under the partial ordering ‘≤’ which is induced by P. For the concepts and properties of the cone, we refer to [1, 2].

We call E a lattice under the partial ordering ‘≤’ if $sup\{x,y\}$ and $inf\{x,y\}$ exist for arbitrary $x,y\in E$. For $x\in E$, let

$$x^{+}=sup\{x,\theta \},x^{-}=inf\{x,\theta \},$$

(1.1)

$x^{+}$ and $x^{-}$ are called the positive part and the negative part of x, respectively, and obviously $x=x^{+}+x^{-}$. Take $|x|=x^{+}-x^{-}$, then $|x|\in P$. One can refer to [3] for the definition and properties of the lattice.

In recent years, many mathematicians have studied a fixed point theorem of nonlinear operators in an ordered Banach space by using topological methods and partial ordered methods (see [4–14] and references therein). However, to the best of our knowledge, few authors have studied the fixed point theorem in Hilbert spaces with lattice structure by applying variational and partial ordered methods. As a result, the goal of this paper is to fill the gap in this area.

Motivated by [7, 8], we obtain some new theorems for nonlinear operators which are not cone mappings by means of variational and partial ordered methods. This paper is organized as follows. Section 2 is devoted to our main results. Section 3 gives examples to indicate the application of our main results.

At the end of this section, we give the following basic concept and lemma from literature which will be used in Section 2.

We consider two real ordered Hilbert spaces in this paper, H with the inner product, norm and cone $(\cdot ,\cdot )$, $\parallel \cdot \parallel$, P, and X with the inner product, norm and cone $P_1$, and we assume that $H\subset X$ and $P\subset P_1$, X is dense in H, the injection being continuous.

Definition 1.1 Let $D\subset H$ and $F:D\to X$ be a nonlinear operator. F is said to be quasi-additive on lattice, if there exists $y\in X$ such that

$$Fx=F{x}^{+}+F{x}^{-}+y,\mathrm{\forall }x\in D,$$

(1.2)

where $x^{+}$ and $x^{-}$ are defined by (1.1). Note that if $F\theta =\theta$, (1.2) becomes $Fx=F{x}^{+}+F{x}^{-}$.

Let $B:X\to H\subset X$ be a bounded linear operator. B is said to be positive if $B(P_1)\subset P$. In this case, B is an increase operator, namely for $x,y\in X$, $x\le y$ implies $Bx\le By$. We have the following conclusion.

Lemma 1.1 ([1, 7])

Suppose that $B:X\to H\subset X$ is a positive bounded linear operator. If the spectral radius $r(B)<1$, then ${(I-B)}^{-1}$ exists and is a positive bounded linear operator. Furthermore,

$${(I-B)}^{-1}=I+B+B^2+\cdots +B^n+\cdots .$$

2 Fixed point theorems

Let H be an ordered real Hilbert space with an ordering given by a closed cone P and suppose that the gradient ${\mathrm{\Phi }}^{\prime }:H\to H$ of a given functional $\mathrm{\Phi }\in C^1(H,R)$ has the expression ${\mathrm{\Phi }}^{\prime }=I-A$. Obviously, the critical points of the functional Φ are the fixed points of the operator A, and vice versa.

We will show that, under additional assumptions on the operator A, Φ satisfies the Palais-Smale compactness condition on a closed convex set $M\subset H$, which ensures the existence of a critical point of Φ, see [15, 16].

• (PS) Every sequence ${\{v_m\}}_{m\in N}\subset M$, satisfying the conditions has a subsequence which converges strongly in M.

  • (i) ${\{\mathrm{\Phi }(v_m)\}}_{m\in N}$ is bounded,

  • (ii) ${lim}_{m\to +\mathrm{\infty }}\parallel {\mathrm{\Phi }}^{\prime }(v_m)\parallel =0$,

The next lemma, the mountain pass lemma [17, 18] on a closed convex subset of a Hilbert space H, is crucial in the proof of our first result.

Lemma 2.1 Assume that H is a Hilbert space, M is a closed convex subset of H, Φ is a $C^1$ functional defined on H, ${\mathrm{\Phi }}^{\prime }(u)$ can be expressed in the form ${\mathrm{\Phi }}^{\prime }(u)=u-Au$, and $A(M)\subset M$. Assume also that Φ satisfies the PS condition on M, Ω is an open subset of M, and there are two points $u_0\in \mathrm{\Omega }$, $u_1\in M\mathrm{\setminus }\mathrm{\Omega }$ such that $max\{\mathrm{\Phi }(u_0),\mathrm{\Phi }(u_1)\}<{inf}_{u\in {\partial }_M\mathrm{\Omega }}\mathrm{\Phi }(u)$. Then

$$c=\underset{h\in {\mathrm{\Phi }}_M}{inf}\underset{t\in [0,1]}{max}f(h(t))$$

is a critical value of Φ and there is at least one critical point in M corresponding to this value, where ${\mathrm{\Phi }}_M=\{h\mid h:[0,1]\to M\mathit{\text{ is continuous, and }}h(0)=u_0,h(1)=u_1\}$.

Theorem 2.1 Let X, H be two ordered real Hilbert spaces. Suppose that $\mathrm{\Phi }\in C^1(H,R)$ satisfies the following hypotheses.

• (i) Φ satisfies the PS condition on H and its gradient ${\mathrm{\Phi }}^{\prime }$ admits the decomposition $I-A$ such that $A=BF$, where $F:H\to X$ is quasi-additive on lattice, $B:X\to H$ is a positive bounded linear operator satisfying:

  • (1) $(Bx,y)\ge 0$ for $\mathrm{\forall }x\in P_1$, $y\in P$;

  • (2) There exists $\phi \in P$ with $\parallel \phi \parallel =1$ such that

    $$B\phi =r(B)\phi ,$$

    that is, φ is the normalized first eigenfunction.

• (ii) There exist $a_1>r^{-1}(B)$ and $y_1\in P_1$ such that

  $$Fx\ge a_{1}x-y_1,x\in P.$$

  (2.1)

• (iii) There exist $0<a_2<r^{-1}(B)$ and $y_2\in P_1$ such that

  $$Fx\ge a_{2}x-y_2,x\in (-P).$$

  (2.2)

• (iv) There exist $M_1\ge M_2>0$ such that $M_2\parallel x\parallel \le {\parallel x\parallel }_1\le M_1\parallel x\parallel$, where ${\parallel x\parallel }_1$ denotes the norm of $|x|$.

• (v) There exist $0<a_3<\frac{1}{M_{1}r(B)}$ and a positive number r such that

  $$|Fx|\le a_3|x|,x\in B_r(\theta ),B_r(\theta )=\{x:\parallel x\parallel \le r\}.$$

Then A has a nontrivial fixed point.

Remark 2.1 From Theorem 3.1 below, we point out that conditions (i) and (iv) of Theorem 2.1 appear naturally in the applications for nonlinear differential equations and integral equations.

Proof of Theorem 2.1 It follows from condition (v) that

$$Fx=F{x}^{+}+F{x}^{-}.$$

Thus, we have

$$Ax=BFx=B(F{x}^{+}+F{x}^{-})=A{x}^{+}+A{x}^{-}.$$

By virtue of (2.1) and (2.2), we have

$$Ax=BFx\ge a_{1}Bx-u_1,x\in P$$

(2.3)

and

$$Ax=BFx\ge a_{2}Bx-u_2,x\in (-P),$$

(2.4)

where $u_1=B{y}_1\in P$ and $u_2=B{y}_2\in P$. Since $0<a_2<r^{-1}(B)$, Lemma 1.1 yields that ${(I-a_{2}B)}^{-1}$ is a positive linear operator. Let $u_0=u_1+u_2$ and $w=-2{(I-a_{2}B)}^{-1}{u}_0$, then $w\in (-P)$ and $w-a_{2}Bw=-2u_0$. This shows $a_{2}Bw-u_0=w+u_0\ge w$. For $x\ge w$, by $w\in (-P)$ we infer $\theta \ge x^{-}\ge w$. On account of (2.3) and (2.4), we arrive at

$$Ax=A{x}^{+}+A{x}^{-}\ge a_{1}B{x}^{+}-u_1+a_{2}B{x}^{-}-u_2\ge a_{1}B{x}^{+}+a_{2}Bw-u_0\ge a_{1}B{x}^{+}+w.$$

This implies

$$A(P_w)\subset P_w,Ax\ge a_{1}Bx+w,\mathrm{\forall }x\in P_w=\{x\in H,x\ge w\}.$$

It is easy to see that $P_w$ is a closed convex subset of H.

By the definition of gradient operator and $\mathrm{\Phi }\in C^1(P_w,R)$, we have

$$\mathrm{\Phi }(x)=\frac{1}{2}{\parallel x\parallel }^2-{\int }_0^1(A(sx),x)ds.$$

Noticing $\mathrm{\Phi }(\theta )=0$, we claim that there is a mountain surrounding θ and $\mathrm{\exists }\alpha >0$, $\mathrm{\exists }\rho >0$ and $x_0\in P_w-B_{\rho }(\theta )$ with $\mathrm{\Phi }(x_0)=0$. Indeed, from (v), we have $r>0$ such that

$$|Fx|\le a_3|x|,x\in B_r(\theta ).$$

Replacing x by $x^{+}$ in the above inequality, we have $|F{x}^{+}|\le a_3{x}^{+}$. Thus,

$$-a_3{x}^{+}\le F{x}^{+}\le a_3{x}^{+},x\in B_r(\theta ).$$

Similarly, we have

$$a_3{x}^{-}\le F{x}^{-}\le -a_3{x}^{-},x\in B_r(\theta ).$$

Thus by (i), for $x\in B_r(\theta )$, we have

$$\begin{array}{rcl}(A(sx),x)& =& (A\left(s{x}^{+}\right)+A\left(s{x}^{-}\right),x^{+}+x^{-})\\ =& (A\left(s{x}^{+}\right),x^{+})+(A\left(s{x}^{+}\right),x^{-})+(A\left(s{x}^{-}\right),x^{+})+(A\left(s{x}^{-}\right),x^{-})\\ \le & (a_{3}B\left(s{x}^{+}\right),x^{+})-(a_{3}B\left(s{x}^{+}\right),x^{-})-(a_{3}B\left(s{x}^{-}\right),x^{+})+(a_{3}B\left(s{x}^{-}\right),x^{-})\\ =& s{a}_3(B(|x|),|x|).\end{array}$$

Let $B_1=a_3{M}_{1}B$. Then by (iv) we get

$$\begin{array}{rcl}\mathrm{\Phi }(x)& \ge & \frac{1}{2}{\parallel x\parallel }^2-{\int }_0^{1}s{a}_3(B|x|,|x|)ds\\ =& \frac{1}{2}({\parallel x\parallel }^2-(a_{3}B|x|,|x|))\\ \ge & \frac{1}{2M_1}({\parallel x\parallel }_1^2-(B_1|x|,|x|))=\frac{1}{2M_1}((I-B_1)|x|,|x|),x\in B_r(\theta ).\end{array}$$

Since $r(B_1)<1$, Lemma 1.1 yields that ${(I-B_1)}^{-1}$ exists and

$${(I-B_1)}^{-1}=I+B_1+B_1^2+\cdots +B_1^n+\cdots .$$

(2.5)

It follows from $B_1(P)\subset P$ that ${(I-B_1)}^{-1}(P)\subset P$. So we know that

$${\parallel x\parallel }_1\le \parallel {(I-B_1)}^{-1}\parallel \parallel y\parallel ,y=(I-B_1)(|x|).$$

Combining with (2.5), we have

$$\begin{array}{rcl}\mathrm{\Phi }(x)& \ge & \frac{1}{2M_1}((I-B_1)(|x|),|x|)=\frac{1}{2M_1}(y,{(I-B_1)}^{-1}y)\\ \ge & \frac{1}{2M_1}(y,y)\ge \frac{1}{2M_1}{\parallel {(I-B_1)}^{-1}\parallel }^{-2}{\parallel x\parallel }_1^2\\ \ge & \frac{M_2^2}{2M_1}{\parallel {(I-B_1)}^{-1}\parallel }^{-2}{\parallel x\parallel }^2,x\in B_r(\theta ).\end{array}$$

Therefore $\mathrm{\exists }\alpha >0$, $\mathrm{\exists }{\rho }^{\prime }>0$ such that $\mathrm{\Phi }{|}_{{\partial }_{P_w}{B}_{{\rho }^{\prime }}(\theta )}\ge \alpha$.

Next, we take $x_0=s\phi$, where $\phi >\theta$ is the normalized first eigenfunction of B, and $s>0$ is to be determined. Set

$$g(s)=\mathrm{\Phi }(s\phi )=\frac{s^2}{2}-{\int }_0^1(A(\tau s\phi ),s\phi )d\tau .$$

From (2.3), we have

$$\begin{array}{rcl}g(s)& \le & \frac{s^2}{2}-{\int }_0^1(a_{1}B(\tau s\phi ),s\phi )d\tau +s\parallel B{y}_1\parallel \\ =& \frac{s^2}{2}(1-(a_{1}B\phi ,\phi ))+s\parallel B{y}_1\parallel \\ =& \frac{s^2}{2}(1-a_{1}r(B))+s\parallel B{y}_1\parallel .\end{array}$$

So we have $g(s)\to -\mathrm{\infty }$ as $s\to +\mathrm{\infty }$. Therefore there exists $s_0>0$ satisfying $g(s_0)=0$, set $s=s_0$, $x_0$ is as required. Hence Theorem 2.1 holds by Lemma 2.1. □

3 Application

Consider the two-point boundary value problem

$$\{\begin{array}{l}{x}^{″}(t)=f(t,x(t)),t\in (0,1),\\ x(0)=x(1)=0.\end{array}$$

(3.1)

Here f is a continuous function from ℝ into ℝ.

Let $H=H_0^1(0,1)$ with the inner product and norm

$$(u,v)={\int }_0^1{u}^{\prime }{v}^{\prime }dt,\parallel u\parallel ={\left({\int }_0^1{\left(u^{\prime }\right)}^{2}dt\right)}^{\frac{1}{2}},$$

and $P=\{u\in H\mid u(t)\ge 0\}$, then H is a lattice under the partial ordering induced by P.

Let $X=L^2(0,1)$ with the cone

$$P_1=\{u\in X\mid u(t)\ge 0\}.$$

For any $x\in H$, it is evident that

$${\left(|x(t)|\right)}^{\prime }=\{\begin{array}{ll}{x}^{\prime }(t)& \text{if }x(t)>0,\\ 0& \text{if }x(t)=0,\\ -x^{\prime }(t)& \text{if }x(t)<0,\end{array}$$

and hence ${\parallel x\parallel }_1=\parallel x\parallel$ and so condition (iv) of Theorem 2.1 holds with $M_1=M_2=1$.

Let

$$(Bx)(t)={\int }_0^{1}G(t,s)x(s)ds$$

(3.2)

with

$$G(t,s)=\{\begin{array}{ll}t(1-s),& 0\le t\le s\le 1,\\ s(1-t),& 0\le s\le t\le 1.\end{array}$$

It is easy to see that $B:P_1\to P$ is a positive bounded linear operator satisfying ${(Bx)}^{″}(t)=-x$. For any $x\in P_1$, $y\in P$, we have

$$\begin{array}{rcl}(Bx,y)& =& {\int }_0^1{(Bx)}^{\prime }(t)y^{\prime }(t)dt\\ =& -{\int }_0^1{(Bx)}^{″}(t)y(t)dt\\ =& {\int }_0^{1}x(t)y(t)dt\ge 0\end{array}$$

and so condition (1) of Theorem 2.1 holds.

As is well known, B has an unbounded sequence of eigenvalues:

$${\lambda }_n=n^2\pi ,n=1,2,\dots .$$

The algebraic multiplicities of every eigenvalue are simple, and the spectral radius $r(B)={\lambda }_1^{-1}=\frac{1}{{\pi }^2}$ and ${\lambda }_{1}B(sin\pi t)=sin\pi t$.

Theorem 3.1 Let $f(t,u):[0,1]\times \mathbb{R}\to \mathbb{R}$ be continuous. Suppose that there exists $\epsilon >0$ such that

$$\underset{u\to -\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{f(t,u)}{u}\le {\lambda }_1-\epsilon ,\mathit{\text{uniformly on }}t\in [0,1],$$

(3.3)

$${\lambda }_1+\epsilon \le \underset{u\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{f(t,u)}{u}\le \underset{u\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{f(t,u)}{u}<+\mathrm{\infty },\mathit{\text{uniformly on }}t\in [0,1],$$

(3.4)

$$\underset{u\to 0}{lim\hspace{0.17em}sup}\left|\frac{f(t,u)}{u}\right|\le {\lambda }_1-\epsilon ,\mathit{\text{uniformly on }}t\in [0,1].$$

(3.5)

Then (3.1) has at least one nontrivial solution.

Proof We define the functional $\mathrm{\Phi }:H\to \mathbb{R}$ as

$$\mathrm{\Phi }(u)={\int }_0^1(\frac{1}{2}{|u^{\prime }(t)|}^2-{\int }_0^{u(t)}f(t,\tau )d\tau )dt.$$

(3.6)

One has that ${\mathrm{\Phi }}^{\prime }(x)y={\int }_0^1{x}^{\prime }{y}^{\prime }dt-{\int }_0^1(Fx)(t)ydt={\int }_0^1(x^{\prime }-{(BFx)}^{\prime }(t))y^{\prime }dt$, which implies that ${\mathrm{\Phi }}^{\prime }(x)=x-BFx$ in H, where B is given by (3.2) and $(Fx)(t)=f(t,x(t))$. Furthermore, from the properties of H, the regularity properties of Φ, it is easy to assert that the solutions to (3.1) are precisely the critical points of Φ.

By means of (3.3), (3.4) and (3.5), we know that there exist $R_0>r>0$ such that

$$\begin{array}{c}f(u)\ge ({\lambda }_1-\frac{\epsilon }{2})u,u\le -R_0,\hfill \\ f(u)\ge ({\lambda }_1+\frac{\epsilon }{2})u,u\ge R_0,\hfill \\ |f(u)|\le ({\lambda }_1-\frac{\epsilon }{2})|u|,|u|\le r,\hfill \end{array}$$

and therefore there exists a constant $M>0$ such that

$$\begin{array}{c}Fx\ge ({\lambda }_1-\frac{\epsilon }{2})x-M,-x\in P,\hfill \\ Fx\ge ({\lambda }_1+\frac{\epsilon }{2})x-M,x\in P,\hfill \\ |Fx|\le ({\lambda }_1-\frac{\epsilon }{2})|x|,\parallel x\parallel \le r,\hfill \end{array}$$

and so conditions (ii), (iii) and (iv) of Theorem 2.1 hold.

From the above discussion, in order to apply Theorem 2.1, we only need to verify that Φ satisfies the PS condition. Suppose that $\{x_n\}\subset H$ satisfies ${\mathrm{\Phi }}^{\prime }(x_n)\to 0$ as $n\to \mathrm{\infty }$ and $|\mathrm{\Phi }(x_n)|\le C$. Taking the inner product of ${\mathrm{\Phi }}^{\prime }(x_n)$ and $x_n^{-}$, we have

$$\circ (1)\parallel x_n^{-}\parallel ={\parallel x_n^{-}\parallel }^2-{\int }_0^{1}f(t,x_n)x_n^{-}dt.$$

We have

$$\begin{array}{rcl}-{\int }_0^{1}f(t,x_n)x_n^{-}dt& \ge & -{\int }_{x(t)\le R_0}f(t,x_n)x_n^{-}dt\\ \ge & -{\int }_{x(t)\le -R_0}f(t,x_n)x_n^{-}dt-C_1\parallel x_n^{-}\parallel ,\end{array}$$

which implies

$$-{\int }_0^{1}f(t,x_n)x_n^{-}dt\ge -({\lambda }_1-\frac{1}{2}\epsilon ){\int }_0^1{\left|x_n^{-}\right|}^{2}dt-C_1\parallel x_n^{-}\parallel .$$

Then we have

$$\circ (1)\parallel x_n^{-}\parallel \ge {\parallel x_n^{-}\parallel }^2-({\lambda }_1-\frac{1}{2}\epsilon ){\int }_0^1{\left|x_n^{-}\right|}^{2}dt-C_1\parallel x_n^{-}\parallel \ge \frac{\epsilon }{2{\lambda }_1}{\parallel x_n^{-}\parallel }^2-C_1\parallel x_n^{-}\parallel ,$$

which gives a bound for $\{x_n^{-}\}$. Then both $\mathrm{\Phi }(x_n^{-})$ and $\mathrm{\Phi }(x_n^{+})$ are bounded. In order to find a bound for $\{x_n^{+}\}$, we use a contradiction argument and assume that $\parallel x_n^{+}\parallel \to \mathrm{\infty }$ as $n\to \mathrm{\infty }$. Defining $v_n=\frac{x_n^{+}}{\parallel x_n^{+}\parallel }$ and selecting a subsequence if necessary, we have $v_n\to v$ weakly in H and strongly in X as $n\to \mathrm{\infty }$ for some $v\in H$, $v\ge 0$. Since ${\int }_0^{u}f(t,\tau )d\tau \le C_2{|u|}^2$ for $u\ge 0$ and $\mathrm{\Phi }(x_n^{+})$ is bounded, it follows from (3.6) that

$$\frac{1}{2}=\frac{\mathrm{\Phi }(x_n^{+})}{{\parallel x_n^{+}\parallel }^2}+{\int }_0^1\frac{{\int }_0^{x_n^{+}}f(t,\tau )d\tau }{{\parallel x_n^{+}\parallel }^2}dt\le \circ (1)+C_2{\int }_0^1{v}_n^{2}dt,$$

which implies that $v\ne 0$. The embedding theorem and the boundedness of $\{x_n^{-}\}$ in H guarantee that there exists $M_1>0$ such that $x_n^{-}\ge -M_1$ for all n. So $f(t,u)$ is bounded for $-M_1\le u\le R_0$. Taking the inner product of $\frac{{\mathrm{\Phi }}^{\prime }(x_n)}{\parallel x_n^{+}\parallel }$ and $sin\pi t$, we see that

$$\begin{array}{rcl}(v_n,sin\pi t)& =& (\frac{{\mathrm{\Phi }}^{\prime }(x_n)}{\parallel x_n^{+}\parallel },sin\pi t)-(\frac{x_n^{-}}{\parallel x_n^{+}\parallel },sin\pi t)+{\int }_0^1\frac{f(t,x_n)}{\parallel x_n^{+}\parallel }sin\pi tdt\\ \ge & \circ (1)+{\int }_0^1\frac{({\lambda }_1+\frac{\epsilon }{2})x_n^{+}-M}{\parallel x_n^{+}\parallel }sin\pi tdt\\ \ge & \circ (1)+({\lambda }_1+\frac{\epsilon }{2}){\int }_0^1{v}_{n}sin\pi tdt.\end{array}$$

Letting $n\to \mathrm{\infty }$, we obtain

$${\lambda }_1{\int }_0^1{v}_{n}sin\pi tdt\ge ({\lambda }_1+\frac{\epsilon }{2}){\int }_0^1{v}_{n}sin\pi tdt,$$

which is a contradiction. Thus, $\{x_n\}$ is bounded in H and it has a convergent subsequence, as a standard consequence. □