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Existence of a fixed point for a semicontinuous operator and its application
Nonlinear Analysis 66 (2007) 1141–1144 www.elsevier.com/locate/na
Existence of a fixed point for a semicontinuous operator and its application Qianqiao Guo ∗ , Pengcheng Niu Department of Applied Mathematics, School of Science, Northwestern Polytechnical University, Xi’an, 710072, People’s Republic of China Received 23 November 2005; accepted 6 January 2006
Abstract We employ a partial order method and cone theory to prove an existence theorem for a fixed point for a semicontinuous operator; also the decreasing and increasing semicontinuous operators are studied. c 2006 Elsevier Ltd. All rights reserved. Keywords: Semicontinuous operator; Fixed point; Cone
1. Introduction Let E be a real Banach space. A cone P is a nonempty closed subset of E with x ∈ P, λ ≥ 0 ⇒ λx ∈ P and x ∈ P, −x ∈ P ⇒ x = θ . The space is ordered by the cone P with x ≤ y(x, y ∈ E) ⇔ y − x ∈ P. A operator A : D → E is decreasing iff x ≤ y implies Ax ≤ Ay; A : D → E is increasing iff x ≤ y implies Ay ≤ Ax; A : D → E is semicontinuous iff x n → x implies Ax n → Ax weakly [5,6]. In nonlinear functional analysis, the problem of establishing the existence and uniqueness of a fixed point for an operator without compactness is difficult, especially for a decreasing or increasing operator. Dajun Guo in [1], Fuyi Li in [2], Zhitao Zhang in [3] and [4], Du in [7] and some others obtained some results on fixed points of decreasing or increasing operators without compactness, but nobody has given the sufficient and necessary conditions. Here we show a result concerning the existence of a fixed point for a semicontinuous operator; as conclusions, ∗ Corresponding author.
E-mail address: [email protected] (Q. Guo). c 2006 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2006.01.010
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we get a sufficient and necessary condition for the existence of a fixed point of a decreasing or increasing operator. 2. Main results Theorem 1. Let E be a real Banach space, P be a cone of E, A : [u, v] → E be semicontinuous, and the following conditions be satisfied: (i) u ≤ Au, Av ≤ v or u ≥ Au, Av ≥ v; (ii) there exists a continuous mapping h : [0, 1] → [u, v], h(0) = u, h(1) = v, such that ∀t ∈ [0, 1], A(h(t)) and h(t) are comparable (i.e. A(h(t)) > h(t) or A(h(t)) ≤ h(t) holds). Then A has at least one fixed point in [u, v]. Proof. We prove this when u ≤ Au, Av ≤ v is satisfied. When u = Au or Av = v or A(h(t0 )) = h(t0 ) for some t0 ∈ [0, 1], the conclusion is obviously true. In the following proof, we suppose u < Au,
Av < v
and A(h(t)) = h(t),
∀t ∈ [0, 1].
(1)
Let g(t) = A(h(t)) − h(t), Δ1 = [0, 1]. l(·) and r (·) denote the left and right borders of an interval respectively. Let t0 = 0, t1 = 1, t2 = 12 ; then g(t0 ) = A(h(t0 )) − h(t0 ) = Au − u > θ, g(t1 ) = A(h(t1 )) − h(t1 ) = Av − v < θ. If g(tn ) > θ
(n = 2, 3, . . .),
let Δn = [tn , r (Δn−1 )], tn+1 =
tn + r (Δn−1 ) ; 2
if g(tn ) < θ
(n = 2, 3, . . .),
let l(Δn−1 ) + tn . 2 It is obvious that g(l(Δn )) > θ, g(r (Δn )) < θ (n = 1, 2, . . .). Then, we get a sequence of closed intervals {Δn } which satisfy Δn = [l(Δn−1 ), tn ], tn+1 =
Δn ⊃ Δn+1
(n = 1, 2, . . .),
lim (r (Δn ) − l(Δn )) = 0.
n→∞
According to the nested interval theorem, there exists a single point ξ ∈ Δn , n = 1, 2, . . ., and lim r (Δn ) = lim l(Δn ) = ξ.
n→∞
n→∞
Q. Guo, P. Niu / Nonlinear Analysis 66 (2007) 1141–1144
1143
Since g is semicontinuous, we have lim g(l(Δn )) ≥ θ,
n→∞
lim g(r (Δn )) ≤ θ.
n→∞
Because of g(ξ ) = lim g(l(Δn )) = lim g(r (Δn )), n→∞
n→∞
we have θ ≤ lim g(l(Δn )) = g(ξ ) = lim g(r (Δn )) ≤ θ. n→∞
n→∞
Hence g(ξ ) = θ , that is A(h(ξ )) = h(ξ ), ξ ∈ Δn ⊂ [0, 1].
(2)
This contradicts (1). Then A has at least one fixed point in [u, v]. The situation is similar for u ≥ Au, Av ≥ v. The proof is finished. In fact, condition (ii) is the sufficient and necessary condition for the existence of a fixed point of a decreasing or increasing operator. Other conditions similar to Theorem 1 are necessary, of course. Theorem 2. If A : [u, v] → E is a semicontinuous and decreasing operator which satisfies u ≤ Au, Av ≤ v, then A has at least one fixed point in [u, v] iff (ii) holds. Proof. We only need to prove that condition (ii) is necessary for the existence of a fixed point for operator A. Suppose x 0 is one fixed point of A in [u, v]. If u < x 0 < v, let ⎧ 1 ⎪ u + 2t (x ⎪ ; − u), t ∈ 0, 0 ⎨ 2 h(t) = ⎪ ⎪ ⎩x 0 + 2 t − 1 (v − x 0 ), t ∈ 1 , 1 . 2 2 Obviously, h : [0, 1] → [u, v] is continuous, h(0) = u, h(1) = v. When t ∈ [0, 12 ], A(h(t)) ≥ Ax 0 = x 0 ≥ h(t); when t ∈ ( 12 , 1], A(h(t)) ≤ Ax 0 = x 0 < h(t). That is, ∀t ∈ [0, 1], A(h(t)) and h(t) are comparable. (ii) holds. If x 0 = u, or x 0 = v, the proof is similar to the above. Theorem 3. If A : [u, v] → E is a semicontinuous and increasing operator which satisfies u ≤ Au, Av ≤ v or u ≥ Au, Av ≥ v, then A has at least one fixed point in [u, v] iff (ii) holds. Proof. The proof is similar to Theorem 2.
Remark. It should be pointed out that the operator does not need to be compact.
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Q. Guo, P. Niu / Nonlinear Analysis 66 (2007) 1141–1144
3. Application Now, we consider the following Hammerstein integral equation:
1 ds = Ax(t). k(t, s) x(t) = N 1 + x(s) R Conclusion. If k(t, s) : R N × R N → R is continuous, where k(t, s) = k1 (t)k2 (s), k1 (t) > 0, k2 (s) > 0, t ∈ R N , s ∈ R N , C B (R N ) denotes the space of bounded and continuous functions, N N N P = C+ B (R ) = {x | x(t) ≥ 0, t ∈ R } is one nonnegative cone in C B (R ), then the ∗ ∗ Hammerstein integral equation has at least one solution x (t) which satisfies x (t) > 0, t ∈ R N . Proof. Obviously, if A : P → P is a continuous and decreasing operator, then A : P → [θ, Aθ ]. It is enough to consider A : [θ, Aθ ] → [θ, Aθ ]. It is easy to see that θ ≤ Aθ, A2 θ ≤ Aθ ;
Aθ = k(t, s)ds = k1 (t) k2 (s)ds.
RN
RN
Let R N k2 (s)ds = a; then Aθ = ak1(t). Let h(r ) = r ak1 (t); we have that h : [0, 1] → [θ, Aθ ] is continuous and
1 ds − r a k1 (t). A(h(r )) − h(r ) = k2 (s) 1 + r ak1(s) RN Since k1 (t) > 0, t ∈ R N , we get that ∀r ∈ [0, 1], Ah(r ) and h(r ) are comparable. (ii) is satisfied. According to Theorem 1, if A has at least one fixed point in [θ, Aθ ], then A has at least one fixed point in P. That is, the Hammerstein integral equation has at least one solution x ∗ (t) which satisfies x ∗ (t) ≥ 0, t ∈ R N . Obviously, x ∗ (t) > 0, t ∈ R N , for k(t, s) = k1 (t)k2 (s), k1 (t) > 0, k2 (s) > 0. Acknowledgement The author thanks Professor Dajun Guo for guidance and encouragement. References [1] D. Guo, A fixed point theorem of decreasing operators and its application, Chinese Sci. Bull. 29 (1984) 189 (in Chinese). [2] F. Li, J. Feng, P. Shen, Fixed point theorem of one kind of decreasing operator and its application, Acta Math. Sinica 42 (2) (1999) 193–196 (in Chinese). [3] Z. Zhang, Some new results about abstract cones and operators, Nonlinear Anal. 37 (1999) 449–455. [4] Z. Zhang, Fixed point theorem of decreasing operator without compactness and its application, J. Systems Sci. Math. Sci. 18 (4) (1998) 422–426 (in Chinese). [5] D.J. Guo, Nonlinear Functional Analysis, 2nd edition, Shandong Sci. and Tec. Press, Jinan, 1985 (in Chinese). [6] D.J. Guo, V. Lakshmikanham, Nonlinear Problems in Abstract Cones, Academic Press, New York, 1988. [7] Y.H. Du, Fixed point of increasing operator in ordered Banach spaces and applications, Appl. Anal. 38 (1990) 1–20.
Existence of a fixed point for a semicontinuous operator and its application Qianqiao Guo ∗ , Pengcheng Niu Department of Applied Mathematics, School of Science, Northwestern Polytechnical University, Xi’an, 710072, People’s Republic of China Received 23 November 2005; accepted 6 January 2006
Abstract We employ a partial order method and cone theory to prove an existence theorem for a fixed point for a semicontinuous operator; also the decreasing and increasing semicontinuous operators are studied. c 2006 Elsevier Ltd. All rights reserved. Keywords: Semicontinuous operator; Fixed point; Cone
1. Introduction Let E be a real Banach space. A cone P is a nonempty closed subset of E with x ∈ P, λ ≥ 0 ⇒ λx ∈ P and x ∈ P, −x ∈ P ⇒ x = θ . The space is ordered by the cone P with x ≤ y(x, y ∈ E) ⇔ y − x ∈ P. A operator A : D → E is decreasing iff x ≤ y implies Ax ≤ Ay; A : D → E is increasing iff x ≤ y implies Ay ≤ Ax; A : D → E is semicontinuous iff x n → x implies Ax n → Ax weakly [5,6]. In nonlinear functional analysis, the problem of establishing the existence and uniqueness of a fixed point for an operator without compactness is difficult, especially for a decreasing or increasing operator. Dajun Guo in [1], Fuyi Li in [2], Zhitao Zhang in [3] and [4], Du in [7] and some others obtained some results on fixed points of decreasing or increasing operators without compactness, but nobody has given the sufficient and necessary conditions. Here we show a result concerning the existence of a fixed point for a semicontinuous operator; as conclusions, ∗ Corresponding author.
E-mail address: [email protected] (Q. Guo). c 2006 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2006.01.010
1142
Q. Guo, P. Niu / Nonlinear Analysis 66 (2007) 1141–1144
we get a sufficient and necessary condition for the existence of a fixed point of a decreasing or increasing operator. 2. Main results Theorem 1. Let E be a real Banach space, P be a cone of E, A : [u, v] → E be semicontinuous, and the following conditions be satisfied: (i) u ≤ Au, Av ≤ v or u ≥ Au, Av ≥ v; (ii) there exists a continuous mapping h : [0, 1] → [u, v], h(0) = u, h(1) = v, such that ∀t ∈ [0, 1], A(h(t)) and h(t) are comparable (i.e. A(h(t)) > h(t) or A(h(t)) ≤ h(t) holds). Then A has at least one fixed point in [u, v]. Proof. We prove this when u ≤ Au, Av ≤ v is satisfied. When u = Au or Av = v or A(h(t0 )) = h(t0 ) for some t0 ∈ [0, 1], the conclusion is obviously true. In the following proof, we suppose u < Au,
Av < v
and A(h(t)) = h(t),
∀t ∈ [0, 1].
(1)
Let g(t) = A(h(t)) − h(t), Δ1 = [0, 1]. l(·) and r (·) denote the left and right borders of an interval respectively. Let t0 = 0, t1 = 1, t2 = 12 ; then g(t0 ) = A(h(t0 )) − h(t0 ) = Au − u > θ, g(t1 ) = A(h(t1 )) − h(t1 ) = Av − v < θ. If g(tn ) > θ
(n = 2, 3, . . .),
let Δn = [tn , r (Δn−1 )], tn+1 =
tn + r (Δn−1 ) ; 2
if g(tn ) < θ
(n = 2, 3, . . .),
let l(Δn−1 ) + tn . 2 It is obvious that g(l(Δn )) > θ, g(r (Δn )) < θ (n = 1, 2, . . .). Then, we get a sequence of closed intervals {Δn } which satisfy Δn = [l(Δn−1 ), tn ], tn+1 =
Δn ⊃ Δn+1
(n = 1, 2, . . .),
lim (r (Δn ) − l(Δn )) = 0.
n→∞
According to the nested interval theorem, there exists a single point ξ ∈ Δn , n = 1, 2, . . ., and lim r (Δn ) = lim l(Δn ) = ξ.
n→∞
n→∞
Q. Guo, P. Niu / Nonlinear Analysis 66 (2007) 1141–1144
1143
Since g is semicontinuous, we have lim g(l(Δn )) ≥ θ,
n→∞
lim g(r (Δn )) ≤ θ.
n→∞
Because of g(ξ ) = lim g(l(Δn )) = lim g(r (Δn )), n→∞
n→∞
we have θ ≤ lim g(l(Δn )) = g(ξ ) = lim g(r (Δn )) ≤ θ. n→∞
n→∞
Hence g(ξ ) = θ , that is A(h(ξ )) = h(ξ ), ξ ∈ Δn ⊂ [0, 1].
(2)
This contradicts (1). Then A has at least one fixed point in [u, v]. The situation is similar for u ≥ Au, Av ≥ v. The proof is finished. In fact, condition (ii) is the sufficient and necessary condition for the existence of a fixed point of a decreasing or increasing operator. Other conditions similar to Theorem 1 are necessary, of course. Theorem 2. If A : [u, v] → E is a semicontinuous and decreasing operator which satisfies u ≤ Au, Av ≤ v, then A has at least one fixed point in [u, v] iff (ii) holds. Proof. We only need to prove that condition (ii) is necessary for the existence of a fixed point for operator A. Suppose x 0 is one fixed point of A in [u, v]. If u < x 0 < v, let ⎧ 1 ⎪ u + 2t (x ⎪ ; − u), t ∈ 0, 0 ⎨ 2 h(t) = ⎪ ⎪ ⎩x 0 + 2 t − 1 (v − x 0 ), t ∈ 1 , 1 . 2 2 Obviously, h : [0, 1] → [u, v] is continuous, h(0) = u, h(1) = v. When t ∈ [0, 12 ], A(h(t)) ≥ Ax 0 = x 0 ≥ h(t); when t ∈ ( 12 , 1], A(h(t)) ≤ Ax 0 = x 0 < h(t). That is, ∀t ∈ [0, 1], A(h(t)) and h(t) are comparable. (ii) holds. If x 0 = u, or x 0 = v, the proof is similar to the above. Theorem 3. If A : [u, v] → E is a semicontinuous and increasing operator which satisfies u ≤ Au, Av ≤ v or u ≥ Au, Av ≥ v, then A has at least one fixed point in [u, v] iff (ii) holds. Proof. The proof is similar to Theorem 2.
Remark. It should be pointed out that the operator does not need to be compact.
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Q. Guo, P. Niu / Nonlinear Analysis 66 (2007) 1141–1144
3. Application Now, we consider the following Hammerstein integral equation:
1 ds = Ax(t). k(t, s) x(t) = N 1 + x(s) R Conclusion. If k(t, s) : R N × R N → R is continuous, where k(t, s) = k1 (t)k2 (s), k1 (t) > 0, k2 (s) > 0, t ∈ R N , s ∈ R N , C B (R N ) denotes the space of bounded and continuous functions, N N N P = C+ B (R ) = {x | x(t) ≥ 0, t ∈ R } is one nonnegative cone in C B (R ), then the ∗ ∗ Hammerstein integral equation has at least one solution x (t) which satisfies x (t) > 0, t ∈ R N . Proof. Obviously, if A : P → P is a continuous and decreasing operator, then A : P → [θ, Aθ ]. It is enough to consider A : [θ, Aθ ] → [θ, Aθ ]. It is easy to see that θ ≤ Aθ, A2 θ ≤ Aθ ;
Aθ = k(t, s)ds = k1 (t) k2 (s)ds.
RN
RN
Let R N k2 (s)ds = a; then Aθ = ak1(t). Let h(r ) = r ak1 (t); we have that h : [0, 1] → [θ, Aθ ] is continuous and
1 ds − r a k1 (t). A(h(r )) − h(r ) = k2 (s) 1 + r ak1(s) RN Since k1 (t) > 0, t ∈ R N , we get that ∀r ∈ [0, 1], Ah(r ) and h(r ) are comparable. (ii) is satisfied. According to Theorem 1, if A has at least one fixed point in [θ, Aθ ], then A has at least one fixed point in P. That is, the Hammerstein integral equation has at least one solution x ∗ (t) which satisfies x ∗ (t) ≥ 0, t ∈ R N . Obviously, x ∗ (t) > 0, t ∈ R N , for k(t, s) = k1 (t)k2 (s), k1 (t) > 0, k2 (s) > 0. Acknowledgement The author thanks Professor Dajun Guo for guidance and encouragement. References [1] D. Guo, A fixed point theorem of decreasing operators and its application, Chinese Sci. Bull. 29 (1984) 189 (in Chinese). [2] F. Li, J. Feng, P. Shen, Fixed point theorem of one kind of decreasing operator and its application, Acta Math. Sinica 42 (2) (1999) 193–196 (in Chinese). [3] Z. Zhang, Some new results about abstract cones and operators, Nonlinear Anal. 37 (1999) 449–455. [4] Z. Zhang, Fixed point theorem of decreasing operator without compactness and its application, J. Systems Sci. Math. Sci. 18 (4) (1998) 422–426 (in Chinese). [5] D.J. Guo, Nonlinear Functional Analysis, 2nd edition, Shandong Sci. and Tec. Press, Jinan, 1985 (in Chinese). [6] D.J. Guo, V. Lakshmikanham, Nonlinear Problems in Abstract Cones, Academic Press, New York, 1988. [7] Y.H. Du, Fixed point of increasing operator in ordered Banach spaces and applications, Appl. Anal. 38 (1990) 1–20.