Corrigendum: On analytic model for two-choice behavior of the paradise fish based on the fixed point method, J. Fixed Po
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Journal of Fixed Point Theory and Applications
Corrigendum: On analytic model for twochoice behavior of the paradise fish based on the fixed point method, J. Fixed Point Theory Appl. 2019, 21:56 Ali Turab and Wutiphol Sintunavarat Abstract. The purpose of this short note is to present some corrections and clarifications concerning the proof of the main result given in the paper “On analytic model for two-choice behavior of the paradise fish based on the fixed point method, J. Fixed Point Theory Appl. 2019, 21:56”. Mathematics Subject Classification. Primary 47H10; Secondary 54H25. Keywords. Choice model, fixed points, Banach fixed point theorem.
1. On the result in [1] In [1], the authors have studied a specific type of choice behavior model for the paradise fish’s learning process. The existence and uniqueness of the solution of the proposed choice behavior model are investigated by using the fixed point tools. We shall first state their results and then discuss some small gaps herewith. In paper [1], the authors denoted by CL0,1 the class of all continuous real-valued functions f : [0, 1] → R such that f (0) = 0 and f (1) = 1, and claimed that CL0,1 , ·CL0,1 is a Banach space, where ·CL0,1 is defined by f CL0,1 = sup x=y
|f (x) − f (y)| |x − y|
(1.1)
for all f ∈ CL0,1 . The authors also proved the following result. Theorem 1.1 (Theorem 5.1 in [1]). For 0 < α ≤ β < 1 with 4β < 1, the mapping T : CL0,1 → CL0,1 defined by (T f )(x) = xf (αx + 1 − α) + (1 − x)f (βx) 0123456789().: V,-vol
(1.2)
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for all f ∈ CL0,1 and x ∈ [0, 1] is a Banach contraction mapping with the metric d induced by ·CL0,1 . We now claim that CL0,1 , ·CL0,1 is not a Banach space with the norm defined above by showing the following counterexample. √ Example 1.2 Let f : [0, 1] → [0, 1] defined by f (x) = x, for all x ∈ [0, 1]. It is easy to see that f is continuous and satisfying the conditions f (0) = 0 and f (1) = 1. Then f ∈ CL0,1 . For each x, y ∈ [0, 1] with x = y, we have √ x − √y |f (x) − f (y)| 1 (1.3) = = √ √ . |x − y| |x − y| x + y Taking x = 0 in (1.3), we get that 1 lim √ = +∞, y
y→0
i.e., the norm defined in (1.1) is not finite. As the zero vector does not belong to CL0,1 , it is not a vector space with condition f (1) = 1.
2. Revised theorem We would like to make the following corrections and clarifications with regard to the text and the proof of the main result: Let us denote by CL0 the class of all continuous real-valued functions f : [0, 1] → R such that f (0) = 0 and |f (x) − f (y)| < ∞. (2.1) |x − y| It is easy to see that CL0 , ·CL0 is a Banach space, where ·CL0 is defined by sup x=y
f CL0 = sup x=y
|f (x) − f (y)| |x − y|
for all f ∈ CL . 0
Theorem 2.1 For 0 < α ≤ β < 1 with 4β < 1, the mapping Z : CL0 → CL0 defined by (Zf )(x) = xf (αx + 1 − α) + (1 − x)f (βx)
(2.2)
for all f ∈ CL0 and x ∈ [0, 1] is a Banach contraction mapping on {f ∈ CL0 | f (1) ≤ 1} with the metric d induced by ·CL0 . To prove this, take the idea of the above co
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