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A Feng Qi Type Inequality for Sugeno Integral

A Feng Qi Type Inequality for Sugeno Integral Fuzzy Inf. Eng. (2010) 3: 293-304 DOI 10.1007/s12543-010-0051-8 ORIGINAL ARTICLE Hamzeh Agahi· Mohammad Ali Yaghoobi Received: 5 April 2010/ Revised: 4 August 2010/ Accepted: 24 August 2010/ © Springer-Verlag Berlin Heidelberg and Fuzzy Information and Engineering Branch of the Operations Research Society of China 2010 Abstract In this paper, a Feng Qi type inequality for Sugeno integral is shown. The studied inequality is based on the classical Feng Qi type inequality for Lebesgue in- tegral. Moreover, a generalized Feng Qi type inequality for Sugeno integral is proved with several examples given to illustrate the validity of the proposed inequalities. Keywords Sugeno integral· Fuzzy measure · Monotone function 1. Introduction In [3], Feng Qi studied a very interesting integral inequality and proved the following result: Theorem 1 Let n be a positive integer. Suppose f (x) has continuous derivative of (i) the n-th order on the interval [a, b] such that f (a) ≥ 0, for 0 ≤ i ≤ n − 1, and (n) f (x) ≥ n!, then n+1 b b n+2 f (x) dx ≥ f (x)dx . (1) a a Some new results on this subject can be found in [4]. In general, any integral inequality can be a very strong tool for applications. In particular, when we think of an integral operator as a predictive tool then an integral inequality can be very important in measuring and dimensioning such process. The concept of Sugeno integral was introduced by Sugeno [11] as a tool for mod- eling nondeterministic problems. Sugeno integral is analogous to Lebesgue integral which has been studied by many authors, including Pap [7], Ralescu and Adams [8], Wang and Klir [12], among others. The difference between Sugeno integral and Lebesgue integral is that addition and multiplication in the definition of Lebesgue in- tegral are replaced respectively by the operations max and min when Sugeno integral is considered. Hamzeh Agahi· Mohammad Ali Yaghoobi () Department of Statistics, Faculty of Mathematics and Computer, Shahid Bahonar University of Kerman, Kerman, Iran email: Yaghoobi@mail.uk.ac.ir 294 Hamzeh Agahi· Mohammad Ali Yaghoobi (2010) Roman-Flores et al [9, 10], started the studies of inequalities for Sugeno integral, and then followed by the authors [1, 2, 5]. In this paper, the Feng Qi type inequality for Sugeno integral is presented with several examples given to illustrate the validity of this inequality. 2. Preliminaries Some definitions and basic properties of the Sugeno integral that will be used in the next sections are as follows. As usual we denote by R and R the set of real and nonnegative real numbers, respectively. Let X be a non-empty set,F be anσ-algebra of subsets of X and m be the Lebesgue measure on R. Throughout this paper, we fix the measurable space (X,F ), and all considered subsets belong to F . Definition 1 (Ralescu and Adams [8]) A set function μ : F→ [0,∞] is called a fuzzy measure if the following properties are satisfied: (FM1)μ(∅) = 0; (FM2) A ⊂ B implies μ(A) ≤ μ(B); (FM3) A ⊂ A ⊂··· implies μ( A ) = lim μ(A ); 1 2 n n→∞ n n=1 (FM4) A ⊃ A ⊃··· , and μ(A ) < +∞ implyμ( A ) = lim μ(A ). 1 2 1 n n→∞ n n=1 Whenμ is a fuzzy measure, the triple (X,F,μ) is called a fuzzy measure space. Let (X,F,μ) be a fuzzy measure space and by F (X) we denote the set of all nonnegative measurable functions f : X −→ [0,∞) with respect to F . In what follows, all considered functions belong to F (X). Let f be a nonnegative real- valued function defined on X. We will denote the set {x ∈ X| f (x) ≥ α} by F for α ≥ 0. Clearly, F is nonincreasing with respect toα, i.e., α ≤ β implies F  F . α α β Definition 2 (Pap [7], Sugeno [11], Wang and Klir [12]) Let (X,F,μ) be a fuzzy measure space and A∈F . Then Sugeno integral of f on A, with respect to the fuzzy measureμ, is defined as (S ) fdμ = (α∧μ(A∩ F )). α≥0 When A = X, then (S ) fdμ = (S ) fdμ = (α∧μ(F )). α≥0 Some basic properties of Sugeno integral are summarized in [7, 12], we cite some of them in the next theorem. Theorem 2 (Pap [7], Wang and Klir [12]) Let (X,F,μ) be a fuzzy measure space. Then i. μ(A∩ F ) ≥ α =⇒ (S ) fdμ ≥ α; ( ) ii. μ(A∩ F ) ≤ α =⇒ S fdμ ≤ α; iii. (S ) fdμ<α ⇐⇒ there existsγ<α such that μ(A∩ F )<α; iv. (S ) fdμ>α ⇐⇒ there existsγ>α such that μ(A∩ F )>α; v. Ifμ(A) < ∞, then μ(A∩ F ) ≤ α ⇐⇒ (S ) fdμ ≤ α; vi. If f ≤ g, then (S ) fdμ ≤ (S ) gdμ. Fuzzy Inf. Eng. (2010) 3: 293-304 295 Roman-Flores et al have studied several fuzzy integral inequalities (see [9, 10]). In particular, the following optimal fuzzy integral inequalities for monotone functions are proved in [9]. Theorem 3 Let μ be the Lebesgue measure on R and let g :[0,∞] → [0,∞] be a continuous and strictly increasing function. If (S ) gdμ = p, then g(a− p) ≥ (S ) gdμ = p, ∀ a ≥ 0. (2) Moreover, if 0 < p < a, then g(a− p) = p. An analogous result is obtained for the decreasing case. Theorem 4 Let μ be the Lebesgue measure on R and let g :[0,∞] → [0,∞] be a continuous and strictly decreasing function. If (S ) gdμ = p, then ( ) g(p) ≥ S gdμ = p, ∀ a ≥ 0. (3) Moreover, if 0 < p < a, then g(p) = p. In [6], Ouyang et al proved the following two theorems which generalized the corresponding results in [9]. Theorem 5 Let m be the Lebesgue measure on R and let g :[0,∞] → [0,∞] be a non-decreasing function. If (S ) gdm = p, then g((a− p)+) ≥ (S ) gdm = p, ∀ a ≥ 0, (4) where g(x+) = lim g(x+ε). Moreover, if p < a, and g is continuous at a− p, then ε−→0 g((a− p)+) = g(a− p) = p. Notice that if m is the Lebesgue measure and g is a non-decreasing function, then g((a− p)+) ≥ p =⇒ (S ) gdm ≥ p. Theorem 6 Let m be the Lebesgue measure on R and let g :[0,∞] → [0,∞] be a non-increasing function. If (S ) gdm = p, then g(p−) ≥ (S ) gdm = p, ∀ a ≥ 0, (5) where g(x−) = lim g(x−ε). Moreover, if p < a, and g is continuous at p, then ε−→0 g(p−) = g(p) = p. Notice that if m is the Lebesgue measure and g is a non-increasing function, then g(p−) ≥ p =⇒ (S ) gdm ≥ p. 3. Main Results The aim of this section is to show Feng Qi type inequality derived from [3] for the Sugeno integral. 296 Hamzeh Agahi· Mohammad Ali Yaghoobi (2010) Theorem 7 Letμ be the Lebesgue measure on R and let f :[0, 1] → [0,∞) be a real- valued function such that (S ) fdμ = p. If f is a continuous and strictly decreasing n+1 n+1 n+2 function, such that f p ≥ p , then the inequality: n+1 1 1 n+2 (S ) f dμ ≥ (S ) fdμ (6) 0 0 holds for all n ≥ 0. 1 1 n+2 Proof Let (S ) f dμ = r and (S ) fdμ = p, where n ≥ 0. Without loss 0 0 of generality, let r ∈ (0, 1) and p ∈ (0, 1]. Then, due to Theorem 4, the following equality holds n+2 f (r) = r. Then n+2 f (r) = r . (7) Now, on the contrary suppose that n+1 r < p . (8) n+1 n+1 n+2 Since f is a continuous and strictly decreasing function such that f p ≥ p and (8) imply that n+1 n+1 n+2 f (r) > f p ≥ p . (9) (7) and (9) imply that 1 n+1 n+1 n+2 n+2 r = f (r) > f p ≥ p . Therefore, n+1 r > p , n+1 which is a contradiction to (8). Hence r ≥ p and the proof is completed. Example 1 Let f be a real valued function defined by f (x) = 1− x where x ∈ [0, 1]. f is a strictly decreasing function. Let n = 1 in (6). A straightforward calculus shows that (i) (S ) f (x)dμ = [α∧μ({1− x ≥ α})] = [α∧ (1−α)] = 0.5. α∈[0,1] α∈[0,1] 3 3 3 (ii) (S ) f (x)dμ = [α∧μ({(1− x) ≥ α})] = [α∧ 1− α ] α∈[0,1] α∈[0,1] = 0. 317 67. Moreover f 0.5 = 0.75 ≥ (0.5) = 0. 629 96. Therefore, 1 1 0. 317 67 = (S ) f dμ ≥ (S ) fdμ = 0.25. 0 0 Fuzzy Inf. Eng. (2010) 3: 293-304 297 n+1 n+1 n+2 Example 2 We observe that f p ≥ p in Theorem 7 cannot be avoided. Let f be a real valued function defined by f (x) = where x ∈ [0, 1]. f is a strictly 1+x decreasing function. Let n = 1. A straightforward calculus shows that 1 1−α (i) (S ) fdμ = α∧μ ≥ α = α∧ = 0. 618 03. α∈[0,1] α∈[0,1] 1+x α 3 1 1 (ii) (S ) f dμ = α∧μ ≥ α = α∧ − 1 α∈[0,1] α∈[0,1] 1+x 0 α = 0. 380 27. Moreover f 0. 618 03 = 0. 723 61  (0. 618 03) = 0. 725 56. Therefore, 1 1 0. 380 27 = (S ) f dμ ≤ (S ) fdμ = 0. 381 96. 0 0 Theorem 8 Letμ be the Lebesgue measure on R and let f :[0, 1] → [0,∞) be a real- valued function such that (S ) fdμ = p. If f is a continuous and strictly increasing n+1 n+1 n+2 function, such that f 1− p ≥ p , then the inequality: n+1 1 1 n+2 (S ) f dμ ≥ (S ) fdμ (10) 0 0 holds for all n ≥ 0. 1 1 n+2 Proof Let (S ) f dμ = r and (S ) fdμ = p, where n ≥ 0. Without loss 0 0 of generality, let r ∈ (0, 1) and p ∈ (0, 1]. Then, due to Theorem 3, the following equality holds n+2 f (1− r) = r. Then n+2 f (1− r) = r . (11) Now, on the contrary suppose that n+1 n+1 r < p =⇒ 1− r > 1− p . (12) n+1 n+1 n+2 Since f is a continuous and strictly increasing function such that f 1− p ≥ p and (12) imply that n+1 n+1 n+2 f (1− r) > f 1− p ≥ p . (13) (11) and (13) imply that 1 n+1 n+1 n+2 n+2 r = f (1− r) > f 1− p ≥ p . Therefore, n+1 r > p , 298 Hamzeh Agahi· Mohammad Ali Yaghoobi (2010) n+1 which is a contradiction to (12). Hence r ≥ p and the proof is completed. Example 3 Let f be a real valued function defined by f (x) = x where x ∈ [0, 1]. f is a strictly increasing function. Let n = 2 in (10). A straightforward calculus shows that (i) (S ) f (x)dμ = [α∧μ({x ≥ α})] = [α∧ (1−α)] = 0.5. α∈[0,1] α∈[0,1] 4 4 (ii) (S ) f (x)dμ = [α∧μ({x ≥ α})] = [α∧ 1− α ] = 0. 275 51. α∈[0,1] α∈[0,1] Moreover f 1− 0.5 = 0. 875 ≥ (0.5) = 0. 629 96. Therefore, 1 1 ( ) ( ) 0. 275 51 = S f dμ ≥ S fdμ = 0.25. 0 0 n+1 n+1 n+2 Example 4 We observe that f 1− p ≥ p in Theorem 8 cannot be avoided. Let f be a real valued function defined by f (x) = where x ∈ [0, 1]. f is a strictly 2−x decreasing function. Let n = 1. A straightforward calculus shows that 1 1 2α− 1 (i) (S ) fdμ = α∧μ ≥ α = α∧ 1− α∈[0,1] α∈[0,1] 2− x α = 0. 618 03. ⎧ ⎫ ⎡ ⎛ ⎞ ⎤ ⎪ ⎪ ⎢ ⎜ ⎪ ⎪ ⎟ ⎥ 1 ⎢ ⎜ 1 ⎟ ⎥ ⎨ ⎬ ⎢ ⎜ ⎟ ⎥ ⎢ ⎜ ⎟ ⎥ (ii) (S ) f dμ = α∧μ ≥ α ⎢ ⎜ ⎪ ⎪ ⎟ ⎥ α∈[0,1] ⎣ ⎝ ⎠ ⎦ 0 ⎪ ⎪ ⎩ ⎭ 2− x 2 α− 1 = α∧ 1− √ α∈[0,1] = 0. 380 27. Moreover f 1− 0. 618 03 = 0. 723 61  (0. 618 03) = 0. 725 56. Therefore, 1 1 0. 380 27 = (S ) f dμ ≤ (S ) fdμ = 0. 381 96. 0 0 4. Further Discussions In this section, the inequalities (6) and (10) are generalized. In fact, the restriction of continuous and strictly increasing (decreasing) are replaced with a more general case as non- decreasing (non-increasing). Furthermore, (S ) (.)dμ is changed to the general from of (S ) (.)dμ, where a ≥ 1. 0 Fuzzy Inf. Eng. (2010) 3: 293-304 299 Theorem 9 Let μ be a Lebesgue measure on [0, a] and f :[0, a] → R be a real- valued measurable function such that (S ) fdμ = p ≤ 1. If f is a non- increasing n+1 n+1 n+2 function such that f p ≥ p , then the inequality: n+1 a a n+2 (S ) f dμ ≥ (S ) fdμ (14) 0 0 holds for all n ≥ 0 and a ≥ 1. a a n+2 ( ) ( ) Proof Let S f dμ = r and S fdμ = p, where n ≥ 0. Without loss 0 0 of generality, let r ∈ (0, 1) and p ∈ (0, 1]. Then, due to Theorem 6, the following equality holds n+2 f (r−) = r. Then n+2 f (r−) = r . (15) Now, on the contrary suppose that n+1 r < p . (16) n+1 n+1 n+2 Since f is a non-increasing function such that f p ≥ p and (16) imply that n+1 n+1 n+2 f (r) ≥ f p ≥ p . (17) (15) and (17) imply that 1 n+1 n+2 n+2 r = f (r−) = lim f (r−ε) ≥ p . ε−→0 Therefore, n+1 r ≥ p , n+1 which is a contradiction to (16). Hence r ≥ p and the proof is completed. Example 5 Let A = [0, 2] and m be the Lebesgue measure. Let f be a real valued function defined by ⎪ x 3 1− , x ∈ [0, ), 2 2 f (x) = 1 3 , x ∈ [ , 2]. 6 2 It is obvious that f is a non- increasing function. Let n = 3 in (14). A straightforward calculus shows that ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ (i) (S ) fdm= α∧ m ≥ α ⎜ ⎟ ⎜ ⎟ ⎝ 6 ⎠ α∈[0, ] ⎛ ⎞ ⎜   ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ α∧ m 1− ≥ α ⎜ ⎟ ⎜ ⎟ ⎝ 2 α∈( ,1] ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎟ ⎟ ⎜ ⎜ ⎜ α∧ ⎟ ⎜ [α∧ (2− 2α)]⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 1 1 α∈[0, ] α∈( ,1] 6 6 = 0. 166 67∨ 0. 666 67 = 0. 666 67. 300 Hamzeh Agahi· Mohammad Ali Yaghoobi (2010) ⎛ ⎞ ⎧ ⎫ ⎡ ⎛ ⎞ ⎤ 2 ⎜ ⎟ ⎪ ⎪ ⎜ ⎢ ⎜ ⎟ ⎥ ⎟ ⎪ ⎪ ⎜ ⎢ ⎜ ⎨ ⎬ ⎟ ⎥ ⎟ 5 ⎜ ⎢ ⎜ ⎟ ⎥ ⎟ ⎜ ⎢ ⎜ ⎟ ⎥ ⎟ (ii) (S ) f dm= ⎜ ⎢ α∧ m ⎜ ≥ α ⎟ ⎥ ⎟ ⎪ ⎪ ⎜ ⎣ ⎝ ⎠ ⎦ ⎟ ⎪ ⎪ ⎝ ⎩ ⎭ ⎠ −4 α∈[0,1. 286×10 ] ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ x ⎟ ⎜ ⎟ ⎜ α∧ m 1− ≥ α ⎟ ⎜ ⎟ ⎝ ⎠ −4 α∈(1. 286×10 ,1] ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 5 ⎟ ⎜ ⎟ ⎜ ⎟ = α∧ α∧ 2− 2 α ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ −4 −4 α∈[0,1. 286×10 ] α∈(1. 286×10 ,1] −4 = 1. 286× 10 ∨ 0. 365 06 = 0. 365 06. In addition f (0. 666 67 ) = 0. 901 23 ≥ 0. 666 67 = 0. 722 98. Therefore, 2 2 0. 365 06 = (S ) f dμ ≥ (S ) fdμ = 0. 197 53. 0 0 n+1 n+1 n+2 Example 6 We observe that f p ≥ p in Theorem 9 cannot be avoided. Let A = [0, 3] and m be the Lebesgue measure. Let f be a real valued function defined by 1− , x ∈ [0, 2.5), f (x) = ⎩ , x ∈ [2.5, 3]. It is obvious that f is a non- increasing function. Let n = 2 in (14). A straightforward calculus shows that ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ (i) (S ) fdm= α∧ m ≥ α ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ α∈[0, ] ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ α∧ m 1− ≥ α ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ α∈( ,1] ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = α∧ [α∧ (8− 8α)] ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ ⎠ 1 1 α∈[0, ] α∈( ,1] 2 2 = 0.5∨ 0. 888 89 = 0. 888 89. ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ (ii) (S ) f dm= α∧ m ≥ α ⎜ ⎟ ⎝ ⎠ α∈[0,0.0 625] ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ α∧ m 1− ≥ α ⎜ ⎟ ⎝ ⎠ α∈(0.0 625,1] ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎜ √ ⎟ ⎜ ⎟ ⎜ 4 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = α∧ α∧ 8− 8 α ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ α∈[0,0.0 625] α∈(0.0 625,1] = 0.0 625∨ 0. 695 12 = 0. 695 12. Fuzzy Inf. Eng. (2010) 3: 293-304 301 Furthermore f (0. 888 89 ) = 0. 912 21  0. 888 89 = 0. 915 45. Therefore, 1 1 ( ) ( ) 0. 695 12 = S f dμ ≤ S fdμ = 0. 702 33. 0 0 Theorem 10 Let μ be a Lebesgue measure on [0, a] and f :[0, a] → R be a real - valued measurable function such that (S ) fdμ = p ≤ 1. If f is a non- decreasing n+1 n+1 n+2 function such that f a− p ≥ p , then the inequality: n+1 a a n+2 (S ) f dμ ≥ (S ) fdμ (18) 0 0 holds for all n ≥ 0 and a ≥ 1. a a n+2 Proof Let (S ) f dμ = r and (S ) fdμ = p, where n ≥ 0. Without loss 0 0 of generality, let r ∈ (0, 1) and p ∈ (0, 1]. Then, due to Theorem 5, the following equality holds n+2 f (a− r+) = r. Then n+2 f (a− r+) = r . (19) Now, on the contrary suppose that n+1 n+1 r < p =⇒ a− r > a− p . (20) n+1 n+1 n+2 Since f is a non-decreasing function such that f a− p ≥ p and (20) imply that n+1 n+1 n+2 f (a− r) ≥ f a− p ≥ p . (21) (19) and (21) imply that 1 n+1 n+2 n+2 r = f (a− r+) = lim f (a− r +ε) ≥ p . ε−→0 Therefore, n+1 r ≥ p , n+1 which is a contradiction to (20). Hence r ≥ p and the proof is completed. Example 7 Let A = [0, 1] and m be the Lebesgue measure. Let f be a real valued function defined by ⎪ 1 1 , x ∈ [0, ], 2 2 f (x) = 2x, x ∈ ( , 1]. 2 302 Hamzeh Agahi· Mohammad Ali Yaghoobi (2010) It is obvious that f is a non- increasing function. Let n = 3 in (18). A straightforward calculus shows that ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ( ) ⎟ (i) (S ) fdm= α∧ m ≥ α [α∧ m {2x ≥ α} ] ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎠ 2 ⎝ 1 1 α∈[0, ] α∈( ,2] 2 2 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜   ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 1 α ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜ α∧ ⎟ ⎜ α∧ 1− ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 1 1 α∈[0, ] α∈( ,2] 2 2 = 0. 5∨ 0. 666 67 = 0. 666 67. ⎛ ⎧ ⎫ ⎞ ⎡ ⎛ ⎞ ⎤ 1 ⎪ ⎪ ⎜ ⎟ ⎢ ⎜ ⎪ ⎪ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ 1 ⎟ ⎥ ⎜ ⎨ ⎬ ⎟ 5 ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ (ii) (S ) f dm= α∧ m ≥ α ⎢ ⎜ ⎪ ⎪ ⎟ ⎥ ⎜ ⎟ ⎣ ⎝ ⎠ ⎦ ⎝ ⎪ ⎪ ⎠ ⎩ ⎭ α∈[0,0.0 312 5] ⎛ ⎞ ⎜    ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ α∧ m (2x) ≥ α ⎜ ⎟ ⎝ ⎠ α∈(0.0 312 5,32] ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ √ ⎟ 1 1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜ α∧ ⎟ ⎜ α∧ 1− α ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 α∈[0,0.0 312 5] α∈(0.0 312 5,32] = 0.0 312 5∨ 0. 555 47 = 0. 555 47. Moreover ) = 1. 604 9 ≥ 0. 666 67 = 0. 722 98. f (1− 0. 666 67 Therefore, 1 1 0. 555 47 = (S ) f dm ≥ (S ) fdm = 0. 197 53. 0 0 n+1 n+1 n+2 Example 8 We observe that f a− p ≥ p in Theorem 10 cannot be avoided. Let A = [0, 1] and m be the Lebesgue measure. Let f be a real valued function defined by 1 1 , x ∈ [0, ], 4 2 f (x) = x 1 , x ∈ ( , 1]. 2 2 It is obvious that f is a non- increasing function. Let n = 1 in (18). A straightforward calculus shows that ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜   ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 1 x ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ (i) (S ) fdm= α∧ m ≥ α α∧ m ≥ α ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 4 ⎠ ⎝ 2 ⎠ 1 1 1 α∈[0, ] α∈( , ] 4 4 2 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ( ) ⎟ = α∧ [α∧ 1− 2α ] ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 1 1 1 α∈[0, ] α∈( , ] 4 4 2 = 0. 25∨ 0.333 33 = 0. 333 33. Fuzzy Inf. Eng. (2010) 3: 293-304 303 ⎛ ⎞ ⎧ ⎫ ⎡ ⎛ ⎞ ⎤ 1 ⎜ ⎟ ⎪ ⎪ ⎜ ⎢ ⎜ ⎟ ⎥ ⎟ ⎪ ⎪ ⎜ ⎢ ⎜ ⎨ ⎬ ⎟ ⎥ ⎟ 3 ⎜ ⎢ ⎜ ⎟ ⎥ ⎟ ⎜ ⎢ ⎜ ⎟ ⎥ ⎟ (ii) (S ) f dm= ⎜ ⎢ α∧ m ⎜ ≥ α ⎟ ⎥ ⎟ ⎪ ⎪ ⎜ ⎣ ⎝ ⎠ ⎦ ⎟ ⎪ ⎪ ⎝ ⎩ ⎭ ⎠ −2 α∈[0,1. 562 5×10 ] ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ x ⎟ ⎜ ⎟ ⎜ α∧ m ≥ α ⎟ ⎜ ⎟ ⎝ ⎠ −2 α∈(1. 562 5×10 ,0. 125] ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟ = α∧ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ −2 α∈[0,1. 562 5×10 ] ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎜ ⎟ α∧ 1− 2 α ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ −2 α∈(1. 562 5×10 ,0. 125] −2 −2 −2 = 1. 562 5× 10 ∨ 9. 320 5× 10 = 9. 320 5× 10 . In addition f (1− 0. 333 33 ) = 0. 444 45  0. 333 33 = 0. 480 75. Therefore, 1 1 −2 3 9. 320 5× 10 = (S ) f dm ≤ (S ) fdm = 0. 111 11. 0 0 5. Conclusion This paper proposed a Feng Qi type inequality for Sugeno integral based on the clas- sical one. First, the usual case of the Sugeno integral over the interval [0, 1] is shown for the Feng Qi type inequality. Secondly, the studied inequality is generalized over the interval [0, a] where a ≥ 1. Furthermore, in the first case, it is assumed that the integrable functions are continuous and strictly increasing (decreasing). However, in the second case, the restrictions on functions are replaced with a more general restric- tion as non-decreasing (non-increasing). Generalization of the above restrictions can be the aim of further research. Some examples are solved to illustrate the proposed inequalities. Acknowledgments This paper is partially supported by the Fuzzy Systems and Applications Center of Excellence, Shahid Bahonar University of Kerman, Kerman, Iran. References 1. Agahi H, Mesiar R, Ouyang Y (2010) General Minkowski type inequalities for Sugeno integrals. Fuzzy Sets and Systems 161: 708-715 2. Agahi H, Mesiar R, Ouyang Y (2009) New general extensions of Chebyshev type inequalities for Sugeno integrals. International Journal of Approximate Reasoning 51: 135-140 3. Qi F (2000) Several integral inequalities. J. Inequal. Pure Appl. Math 1(2) Art 19. Available online at http://jipam.vu.edu.au/volumes.php 4. Yu K W, Qi F (2001) A short note on an integral inequality. RGMIA Res. Rep. Coll. 4(1): 23-25 5. Mesiar R, Ouyang Y (2009) General Chebyshev type inequalities for Sugeno integrals, Fuzzy Sets and Systems 160: 58-64 304 Hamzeh Agahi· Mohammad Ali Yaghoobi (2010) 6. Ouyang Y, Fang J (2008) Sugeno integral of monotone functions based on Lebesgue measure. Com- puters and Mathematics with Applications 56: 367-374 7. Pap E (1995) Null-additive set functions. Kluwer, Dordrecht 8. Ralescu D, Adams G (1980) The fuzzy integral. Journal of Mathematical Analysis and Applications 75: 562-570 9. Roman-Flores ´ H, Flores-Franulic ˇ A, Chalco-Cano Y (2007) The fuzzy integral for monotone func- tions. Applied Mathematics and Computation 185: 492-498 10. Roman-Flores ´ H, Flores-Franulic ˇ A, Chalco-Cano Y (2007) A Jensen type inequality for fuzzy inte- grals. Information Sciences 177: 3192-3201 11. Sugeno M (1974) Theory of fuzzy integrals and its applications. Ph.D thesis, Tokyo Institute of Technology 12. Wang Z, Klir G (1992) Fuzzy measure theory. Plenum Press, New York http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Fuzzy Information and Engineering Taylor & Francis

A Feng Qi Type Inequality for Sugeno Integral

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Fuzzy Inf. Eng. (2010) 3: 293-304 DOI 10.1007/s12543-010-0051-8 ORIGINAL ARTICLE Hamzeh Agahi· Mohammad Ali Yaghoobi Received: 5 April 2010/ Revised: 4 August 2010/ Accepted: 24 August 2010/ © Springer-Verlag Berlin Heidelberg and Fuzzy Information and Engineering Branch of the Operations Research Society of China 2010 Abstract In this paper, a Feng Qi type inequality for Sugeno integral is shown. The studied inequality is based on the classical Feng Qi type inequality for Lebesgue in- tegral. Moreover, a generalized Feng Qi type inequality for Sugeno integral is proved with several examples given to illustrate the validity of the proposed inequalities. Keywords Sugeno integral· Fuzzy measure · Monotone function 1. Introduction In [3], Feng Qi studied a very interesting integral inequality and proved the following result: Theorem 1 Let n be a positive integer. Suppose f (x) has continuous derivative of (i) the n-th order on the interval [a, b] such that f (a) ≥ 0, for 0 ≤ i ≤ n − 1, and (n) f (x) ≥ n!, then n+1 b b n+2 f (x) dx ≥ f (x)dx . (1) a a Some new results on this subject can be found in [4]. In general, any integral inequality can be a very strong tool for applications. In particular, when we think of an integral operator as a predictive tool then an integral inequality can be very important in measuring and dimensioning such process. The concept of Sugeno integral was introduced by Sugeno [11] as a tool for mod- eling nondeterministic problems. Sugeno integral is analogous to Lebesgue integral which has been studied by many authors, including Pap [7], Ralescu and Adams [8], Wang and Klir [12], among others. The difference between Sugeno integral and Lebesgue integral is that addition and multiplication in the definition of Lebesgue in- tegral are replaced respectively by the operations max and min when Sugeno integral is considered. Hamzeh Agahi· Mohammad Ali Yaghoobi () Department of Statistics, Faculty of Mathematics and Computer, Shahid Bahonar University of Kerman, Kerman, Iran email: Yaghoobi@mail.uk.ac.ir 294 Hamzeh Agahi· Mohammad Ali Yaghoobi (2010) Roman-Flores et al [9, 10], started the studies of inequalities for Sugeno integral, and then followed by the authors [1, 2, 5]. In this paper, the Feng Qi type inequality for Sugeno integral is presented with several examples given to illustrate the validity of this inequality. 2. Preliminaries Some definitions and basic properties of the Sugeno integral that will be used in the next sections are as follows. As usual we denote by R and R the set of real and nonnegative real numbers, respectively. Let X be a non-empty set,F be anσ-algebra of subsets of X and m be the Lebesgue measure on R. Throughout this paper, we fix the measurable space (X,F ), and all considered subsets belong to F . Definition 1 (Ralescu and Adams [8]) A set function μ : F→ [0,∞] is called a fuzzy measure if the following properties are satisfied: (FM1)μ(∅) = 0; (FM2) A ⊂ B implies μ(A) ≤ μ(B); (FM3) A ⊂ A ⊂··· implies μ( A ) = lim μ(A ); 1 2 n n→∞ n n=1 (FM4) A ⊃ A ⊃··· , and μ(A ) < +∞ implyμ( A ) = lim μ(A ). 1 2 1 n n→∞ n n=1 Whenμ is a fuzzy measure, the triple (X,F,μ) is called a fuzzy measure space. Let (X,F,μ) be a fuzzy measure space and by F (X) we denote the set of all nonnegative measurable functions f : X −→ [0,∞) with respect to F . In what follows, all considered functions belong to F (X). Let f be a nonnegative real- valued function defined on X. We will denote the set {x ∈ X| f (x) ≥ α} by F for α ≥ 0. Clearly, F is nonincreasing with respect toα, i.e., α ≤ β implies F  F . α α β Definition 2 (Pap [7], Sugeno [11], Wang and Klir [12]) Let (X,F,μ) be a fuzzy measure space and A∈F . Then Sugeno integral of f on A, with respect to the fuzzy measureμ, is defined as (S ) fdμ = (α∧μ(A∩ F )). α≥0 When A = X, then (S ) fdμ = (S ) fdμ = (α∧μ(F )). α≥0 Some basic properties of Sugeno integral are summarized in [7, 12], we cite some of them in the next theorem. Theorem 2 (Pap [7], Wang and Klir [12]) Let (X,F,μ) be a fuzzy measure space. Then i. μ(A∩ F ) ≥ α =⇒ (S ) fdμ ≥ α; ( ) ii. μ(A∩ F ) ≤ α =⇒ S fdμ ≤ α; iii. (S ) fdμ<α ⇐⇒ there existsγ<α such that μ(A∩ F )<α; iv. (S ) fdμ>α ⇐⇒ there existsγ>α such that μ(A∩ F )>α; v. Ifμ(A) < ∞, then μ(A∩ F ) ≤ α ⇐⇒ (S ) fdμ ≤ α; vi. If f ≤ g, then (S ) fdμ ≤ (S ) gdμ. Fuzzy Inf. Eng. (2010) 3: 293-304 295 Roman-Flores et al have studied several fuzzy integral inequalities (see [9, 10]). In particular, the following optimal fuzzy integral inequalities for monotone functions are proved in [9]. Theorem 3 Let μ be the Lebesgue measure on R and let g :[0,∞] → [0,∞] be a continuous and strictly increasing function. If (S ) gdμ = p, then g(a− p) ≥ (S ) gdμ = p, ∀ a ≥ 0. (2) Moreover, if 0 < p < a, then g(a− p) = p. An analogous result is obtained for the decreasing case. Theorem 4 Let μ be the Lebesgue measure on R and let g :[0,∞] → [0,∞] be a continuous and strictly decreasing function. If (S ) gdμ = p, then ( ) g(p) ≥ S gdμ = p, ∀ a ≥ 0. (3) Moreover, if 0 < p < a, then g(p) = p. In [6], Ouyang et al proved the following two theorems which generalized the corresponding results in [9]. Theorem 5 Let m be the Lebesgue measure on R and let g :[0,∞] → [0,∞] be a non-decreasing function. If (S ) gdm = p, then g((a− p)+) ≥ (S ) gdm = p, ∀ a ≥ 0, (4) where g(x+) = lim g(x+ε). Moreover, if p < a, and g is continuous at a− p, then ε−→0 g((a− p)+) = g(a− p) = p. Notice that if m is the Lebesgue measure and g is a non-decreasing function, then g((a− p)+) ≥ p =⇒ (S ) gdm ≥ p. Theorem 6 Let m be the Lebesgue measure on R and let g :[0,∞] → [0,∞] be a non-increasing function. If (S ) gdm = p, then g(p−) ≥ (S ) gdm = p, ∀ a ≥ 0, (5) where g(x−) = lim g(x−ε). Moreover, if p < a, and g is continuous at p, then ε−→0 g(p−) = g(p) = p. Notice that if m is the Lebesgue measure and g is a non-increasing function, then g(p−) ≥ p =⇒ (S ) gdm ≥ p. 3. Main Results The aim of this section is to show Feng Qi type inequality derived from [3] for the Sugeno integral. 296 Hamzeh Agahi· Mohammad Ali Yaghoobi (2010) Theorem 7 Letμ be the Lebesgue measure on R and let f :[0, 1] → [0,∞) be a real- valued function such that (S ) fdμ = p. If f is a continuous and strictly decreasing n+1 n+1 n+2 function, such that f p ≥ p , then the inequality: n+1 1 1 n+2 (S ) f dμ ≥ (S ) fdμ (6) 0 0 holds for all n ≥ 0. 1 1 n+2 Proof Let (S ) f dμ = r and (S ) fdμ = p, where n ≥ 0. Without loss 0 0 of generality, let r ∈ (0, 1) and p ∈ (0, 1]. Then, due to Theorem 4, the following equality holds n+2 f (r) = r. Then n+2 f (r) = r . (7) Now, on the contrary suppose that n+1 r < p . (8) n+1 n+1 n+2 Since f is a continuous and strictly decreasing function such that f p ≥ p and (8) imply that n+1 n+1 n+2 f (r) > f p ≥ p . (9) (7) and (9) imply that 1 n+1 n+1 n+2 n+2 r = f (r) > f p ≥ p . Therefore, n+1 r > p , n+1 which is a contradiction to (8). Hence r ≥ p and the proof is completed. Example 1 Let f be a real valued function defined by f (x) = 1− x where x ∈ [0, 1]. f is a strictly decreasing function. Let n = 1 in (6). A straightforward calculus shows that (i) (S ) f (x)dμ = [α∧μ({1− x ≥ α})] = [α∧ (1−α)] = 0.5. α∈[0,1] α∈[0,1] 3 3 3 (ii) (S ) f (x)dμ = [α∧μ({(1− x) ≥ α})] = [α∧ 1− α ] α∈[0,1] α∈[0,1] = 0. 317 67. Moreover f 0.5 = 0.75 ≥ (0.5) = 0. 629 96. Therefore, 1 1 0. 317 67 = (S ) f dμ ≥ (S ) fdμ = 0.25. 0 0 Fuzzy Inf. Eng. (2010) 3: 293-304 297 n+1 n+1 n+2 Example 2 We observe that f p ≥ p in Theorem 7 cannot be avoided. Let f be a real valued function defined by f (x) = where x ∈ [0, 1]. f is a strictly 1+x decreasing function. Let n = 1. A straightforward calculus shows that 1 1−α (i) (S ) fdμ = α∧μ ≥ α = α∧ = 0. 618 03. α∈[0,1] α∈[0,1] 1+x α 3 1 1 (ii) (S ) f dμ = α∧μ ≥ α = α∧ − 1 α∈[0,1] α∈[0,1] 1+x 0 α = 0. 380 27. Moreover f 0. 618 03 = 0. 723 61  (0. 618 03) = 0. 725 56. Therefore, 1 1 0. 380 27 = (S ) f dμ ≤ (S ) fdμ = 0. 381 96. 0 0 Theorem 8 Letμ be the Lebesgue measure on R and let f :[0, 1] → [0,∞) be a real- valued function such that (S ) fdμ = p. If f is a continuous and strictly increasing n+1 n+1 n+2 function, such that f 1− p ≥ p , then the inequality: n+1 1 1 n+2 (S ) f dμ ≥ (S ) fdμ (10) 0 0 holds for all n ≥ 0. 1 1 n+2 Proof Let (S ) f dμ = r and (S ) fdμ = p, where n ≥ 0. Without loss 0 0 of generality, let r ∈ (0, 1) and p ∈ (0, 1]. Then, due to Theorem 3, the following equality holds n+2 f (1− r) = r. Then n+2 f (1− r) = r . (11) Now, on the contrary suppose that n+1 n+1 r < p =⇒ 1− r > 1− p . (12) n+1 n+1 n+2 Since f is a continuous and strictly increasing function such that f 1− p ≥ p and (12) imply that n+1 n+1 n+2 f (1− r) > f 1− p ≥ p . (13) (11) and (13) imply that 1 n+1 n+1 n+2 n+2 r = f (1− r) > f 1− p ≥ p . Therefore, n+1 r > p , 298 Hamzeh Agahi· Mohammad Ali Yaghoobi (2010) n+1 which is a contradiction to (12). Hence r ≥ p and the proof is completed. Example 3 Let f be a real valued function defined by f (x) = x where x ∈ [0, 1]. f is a strictly increasing function. Let n = 2 in (10). A straightforward calculus shows that (i) (S ) f (x)dμ = [α∧μ({x ≥ α})] = [α∧ (1−α)] = 0.5. α∈[0,1] α∈[0,1] 4 4 (ii) (S ) f (x)dμ = [α∧μ({x ≥ α})] = [α∧ 1− α ] = 0. 275 51. α∈[0,1] α∈[0,1] Moreover f 1− 0.5 = 0. 875 ≥ (0.5) = 0. 629 96. Therefore, 1 1 ( ) ( ) 0. 275 51 = S f dμ ≥ S fdμ = 0.25. 0 0 n+1 n+1 n+2 Example 4 We observe that f 1− p ≥ p in Theorem 8 cannot be avoided. Let f be a real valued function defined by f (x) = where x ∈ [0, 1]. f is a strictly 2−x decreasing function. Let n = 1. A straightforward calculus shows that 1 1 2α− 1 (i) (S ) fdμ = α∧μ ≥ α = α∧ 1− α∈[0,1] α∈[0,1] 2− x α = 0. 618 03. ⎧ ⎫ ⎡ ⎛ ⎞ ⎤ ⎪ ⎪ ⎢ ⎜ ⎪ ⎪ ⎟ ⎥ 1 ⎢ ⎜ 1 ⎟ ⎥ ⎨ ⎬ ⎢ ⎜ ⎟ ⎥ ⎢ ⎜ ⎟ ⎥ (ii) (S ) f dμ = α∧μ ≥ α ⎢ ⎜ ⎪ ⎪ ⎟ ⎥ α∈[0,1] ⎣ ⎝ ⎠ ⎦ 0 ⎪ ⎪ ⎩ ⎭ 2− x 2 α− 1 = α∧ 1− √ α∈[0,1] = 0. 380 27. Moreover f 1− 0. 618 03 = 0. 723 61  (0. 618 03) = 0. 725 56. Therefore, 1 1 0. 380 27 = (S ) f dμ ≤ (S ) fdμ = 0. 381 96. 0 0 4. Further Discussions In this section, the inequalities (6) and (10) are generalized. In fact, the restriction of continuous and strictly increasing (decreasing) are replaced with a more general case as non- decreasing (non-increasing). Furthermore, (S ) (.)dμ is changed to the general from of (S ) (.)dμ, where a ≥ 1. 0 Fuzzy Inf. Eng. (2010) 3: 293-304 299 Theorem 9 Let μ be a Lebesgue measure on [0, a] and f :[0, a] → R be a real- valued measurable function such that (S ) fdμ = p ≤ 1. If f is a non- increasing n+1 n+1 n+2 function such that f p ≥ p , then the inequality: n+1 a a n+2 (S ) f dμ ≥ (S ) fdμ (14) 0 0 holds for all n ≥ 0 and a ≥ 1. a a n+2 ( ) ( ) Proof Let S f dμ = r and S fdμ = p, where n ≥ 0. Without loss 0 0 of generality, let r ∈ (0, 1) and p ∈ (0, 1]. Then, due to Theorem 6, the following equality holds n+2 f (r−) = r. Then n+2 f (r−) = r . (15) Now, on the contrary suppose that n+1 r < p . (16) n+1 n+1 n+2 Since f is a non-increasing function such that f p ≥ p and (16) imply that n+1 n+1 n+2 f (r) ≥ f p ≥ p . (17) (15) and (17) imply that 1 n+1 n+2 n+2 r = f (r−) = lim f (r−ε) ≥ p . ε−→0 Therefore, n+1 r ≥ p , n+1 which is a contradiction to (16). Hence r ≥ p and the proof is completed. Example 5 Let A = [0, 2] and m be the Lebesgue measure. Let f be a real valued function defined by ⎪ x 3 1− , x ∈ [0, ), 2 2 f (x) = 1 3 , x ∈ [ , 2]. 6 2 It is obvious that f is a non- increasing function. Let n = 3 in (14). A straightforward calculus shows that ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ (i) (S ) fdm= α∧ m ≥ α ⎜ ⎟ ⎜ ⎟ ⎝ 6 ⎠ α∈[0, ] ⎛ ⎞ ⎜   ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ α∧ m 1− ≥ α ⎜ ⎟ ⎜ ⎟ ⎝ 2 α∈( ,1] ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎟ ⎟ ⎜ ⎜ ⎜ α∧ ⎟ ⎜ [α∧ (2− 2α)]⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 1 1 α∈[0, ] α∈( ,1] 6 6 = 0. 166 67∨ 0. 666 67 = 0. 666 67. 300 Hamzeh Agahi· Mohammad Ali Yaghoobi (2010) ⎛ ⎞ ⎧ ⎫ ⎡ ⎛ ⎞ ⎤ 2 ⎜ ⎟ ⎪ ⎪ ⎜ ⎢ ⎜ ⎟ ⎥ ⎟ ⎪ ⎪ ⎜ ⎢ ⎜ ⎨ ⎬ ⎟ ⎥ ⎟ 5 ⎜ ⎢ ⎜ ⎟ ⎥ ⎟ ⎜ ⎢ ⎜ ⎟ ⎥ ⎟ (ii) (S ) f dm= ⎜ ⎢ α∧ m ⎜ ≥ α ⎟ ⎥ ⎟ ⎪ ⎪ ⎜ ⎣ ⎝ ⎠ ⎦ ⎟ ⎪ ⎪ ⎝ ⎩ ⎭ ⎠ −4 α∈[0,1. 286×10 ] ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ x ⎟ ⎜ ⎟ ⎜ α∧ m 1− ≥ α ⎟ ⎜ ⎟ ⎝ ⎠ −4 α∈(1. 286×10 ,1] ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 5 ⎟ ⎜ ⎟ ⎜ ⎟ = α∧ α∧ 2− 2 α ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ −4 −4 α∈[0,1. 286×10 ] α∈(1. 286×10 ,1] −4 = 1. 286× 10 ∨ 0. 365 06 = 0. 365 06. In addition f (0. 666 67 ) = 0. 901 23 ≥ 0. 666 67 = 0. 722 98. Therefore, 2 2 0. 365 06 = (S ) f dμ ≥ (S ) fdμ = 0. 197 53. 0 0 n+1 n+1 n+2 Example 6 We observe that f p ≥ p in Theorem 9 cannot be avoided. Let A = [0, 3] and m be the Lebesgue measure. Let f be a real valued function defined by 1− , x ∈ [0, 2.5), f (x) = ⎩ , x ∈ [2.5, 3]. It is obvious that f is a non- increasing function. Let n = 2 in (14). A straightforward calculus shows that ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ (i) (S ) fdm= α∧ m ≥ α ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ α∈[0, ] ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ α∧ m 1− ≥ α ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ α∈( ,1] ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = α∧ [α∧ (8− 8α)] ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ ⎠ 1 1 α∈[0, ] α∈( ,1] 2 2 = 0.5∨ 0. 888 89 = 0. 888 89. ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ (ii) (S ) f dm= α∧ m ≥ α ⎜ ⎟ ⎝ ⎠ α∈[0,0.0 625] ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ α∧ m 1− ≥ α ⎜ ⎟ ⎝ ⎠ α∈(0.0 625,1] ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎜ √ ⎟ ⎜ ⎟ ⎜ 4 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = α∧ α∧ 8− 8 α ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ α∈[0,0.0 625] α∈(0.0 625,1] = 0.0 625∨ 0. 695 12 = 0. 695 12. Fuzzy Inf. Eng. (2010) 3: 293-304 301 Furthermore f (0. 888 89 ) = 0. 912 21  0. 888 89 = 0. 915 45. Therefore, 1 1 ( ) ( ) 0. 695 12 = S f dμ ≤ S fdμ = 0. 702 33. 0 0 Theorem 10 Let μ be a Lebesgue measure on [0, a] and f :[0, a] → R be a real - valued measurable function such that (S ) fdμ = p ≤ 1. If f is a non- decreasing n+1 n+1 n+2 function such that f a− p ≥ p , then the inequality: n+1 a a n+2 (S ) f dμ ≥ (S ) fdμ (18) 0 0 holds for all n ≥ 0 and a ≥ 1. a a n+2 Proof Let (S ) f dμ = r and (S ) fdμ = p, where n ≥ 0. Without loss 0 0 of generality, let r ∈ (0, 1) and p ∈ (0, 1]. Then, due to Theorem 5, the following equality holds n+2 f (a− r+) = r. Then n+2 f (a− r+) = r . (19) Now, on the contrary suppose that n+1 n+1 r < p =⇒ a− r > a− p . (20) n+1 n+1 n+2 Since f is a non-decreasing function such that f a− p ≥ p and (20) imply that n+1 n+1 n+2 f (a− r) ≥ f a− p ≥ p . (21) (19) and (21) imply that 1 n+1 n+2 n+2 r = f (a− r+) = lim f (a− r +ε) ≥ p . ε−→0 Therefore, n+1 r ≥ p , n+1 which is a contradiction to (20). Hence r ≥ p and the proof is completed. Example 7 Let A = [0, 1] and m be the Lebesgue measure. Let f be a real valued function defined by ⎪ 1 1 , x ∈ [0, ], 2 2 f (x) = 2x, x ∈ ( , 1]. 2 302 Hamzeh Agahi· Mohammad Ali Yaghoobi (2010) It is obvious that f is a non- increasing function. Let n = 3 in (18). A straightforward calculus shows that ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ( ) ⎟ (i) (S ) fdm= α∧ m ≥ α [α∧ m {2x ≥ α} ] ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎠ 2 ⎝ 1 1 α∈[0, ] α∈( ,2] 2 2 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜   ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 1 α ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜ α∧ ⎟ ⎜ α∧ 1− ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 1 1 α∈[0, ] α∈( ,2] 2 2 = 0. 5∨ 0. 666 67 = 0. 666 67. ⎛ ⎧ ⎫ ⎞ ⎡ ⎛ ⎞ ⎤ 1 ⎪ ⎪ ⎜ ⎟ ⎢ ⎜ ⎪ ⎪ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ 1 ⎟ ⎥ ⎜ ⎨ ⎬ ⎟ 5 ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ (ii) (S ) f dm= α∧ m ≥ α ⎢ ⎜ ⎪ ⎪ ⎟ ⎥ ⎜ ⎟ ⎣ ⎝ ⎠ ⎦ ⎝ ⎪ ⎪ ⎠ ⎩ ⎭ α∈[0,0.0 312 5] ⎛ ⎞ ⎜    ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ α∧ m (2x) ≥ α ⎜ ⎟ ⎝ ⎠ α∈(0.0 312 5,32] ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ √ ⎟ 1 1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜ α∧ ⎟ ⎜ α∧ 1− α ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 α∈[0,0.0 312 5] α∈(0.0 312 5,32] = 0.0 312 5∨ 0. 555 47 = 0. 555 47. Moreover ) = 1. 604 9 ≥ 0. 666 67 = 0. 722 98. f (1− 0. 666 67 Therefore, 1 1 0. 555 47 = (S ) f dm ≥ (S ) fdm = 0. 197 53. 0 0 n+1 n+1 n+2 Example 8 We observe that f a− p ≥ p in Theorem 10 cannot be avoided. Let A = [0, 1] and m be the Lebesgue measure. Let f be a real valued function defined by 1 1 , x ∈ [0, ], 4 2 f (x) = x 1 , x ∈ ( , 1]. 2 2 It is obvious that f is a non- increasing function. Let n = 1 in (18). A straightforward calculus shows that ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜   ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 1 x ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ (i) (S ) fdm= α∧ m ≥ α α∧ m ≥ α ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 4 ⎠ ⎝ 2 ⎠ 1 1 1 α∈[0, ] α∈( , ] 4 4 2 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ( ) ⎟ = α∧ [α∧ 1− 2α ] ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 1 1 1 α∈[0, ] α∈( , ] 4 4 2 = 0. 25∨ 0.333 33 = 0. 333 33. Fuzzy Inf. Eng. (2010) 3: 293-304 303 ⎛ ⎞ ⎧ ⎫ ⎡ ⎛ ⎞ ⎤ 1 ⎜ ⎟ ⎪ ⎪ ⎜ ⎢ ⎜ ⎟ ⎥ ⎟ ⎪ ⎪ ⎜ ⎢ ⎜ ⎨ ⎬ ⎟ ⎥ ⎟ 3 ⎜ ⎢ ⎜ ⎟ ⎥ ⎟ ⎜ ⎢ ⎜ ⎟ ⎥ ⎟ (ii) (S ) f dm= ⎜ ⎢ α∧ m ⎜ ≥ α ⎟ ⎥ ⎟ ⎪ ⎪ ⎜ ⎣ ⎝ ⎠ ⎦ ⎟ ⎪ ⎪ ⎝ ⎩ ⎭ ⎠ −2 α∈[0,1. 562 5×10 ] ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ x ⎟ ⎜ ⎟ ⎜ α∧ m ≥ α ⎟ ⎜ ⎟ ⎝ ⎠ −2 α∈(1. 562 5×10 ,0. 125] ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟ = α∧ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ −2 α∈[0,1. 562 5×10 ] ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎜ ⎟ α∧ 1− 2 α ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ −2 α∈(1. 562 5×10 ,0. 125] −2 −2 −2 = 1. 562 5× 10 ∨ 9. 320 5× 10 = 9. 320 5× 10 . In addition f (1− 0. 333 33 ) = 0. 444 45  0. 333 33 = 0. 480 75. Therefore, 1 1 −2 3 9. 320 5× 10 = (S ) f dm ≤ (S ) fdm = 0. 111 11. 0 0 5. Conclusion This paper proposed a Feng Qi type inequality for Sugeno integral based on the clas- sical one. First, the usual case of the Sugeno integral over the interval [0, 1] is shown for the Feng Qi type inequality. Secondly, the studied inequality is generalized over the interval [0, a] where a ≥ 1. Furthermore, in the first case, it is assumed that the integrable functions are continuous and strictly increasing (decreasing). However, in the second case, the restrictions on functions are replaced with a more general restric- tion as non-decreasing (non-increasing). Generalization of the above restrictions can be the aim of further research. Some examples are solved to illustrate the proposed inequalities. Acknowledgments This paper is partially supported by the Fuzzy Systems and Applications Center of Excellence, Shahid Bahonar University of Kerman, Kerman, Iran. References 1. Agahi H, Mesiar R, Ouyang Y (2010) General Minkowski type inequalities for Sugeno integrals. Fuzzy Sets and Systems 161: 708-715 2. Agahi H, Mesiar R, Ouyang Y (2009) New general extensions of Chebyshev type inequalities for Sugeno integrals. International Journal of Approximate Reasoning 51: 135-140 3. Qi F (2000) Several integral inequalities. J. Inequal. Pure Appl. Math 1(2) Art 19. Available online at http://jipam.vu.edu.au/volumes.php 4. Yu K W, Qi F (2001) A short note on an integral inequality. RGMIA Res. Rep. Coll. 4(1): 23-25 5. Mesiar R, Ouyang Y (2009) General Chebyshev type inequalities for Sugeno integrals, Fuzzy Sets and Systems 160: 58-64 304 Hamzeh Agahi· Mohammad Ali Yaghoobi (2010) 6. Ouyang Y, Fang J (2008) Sugeno integral of monotone functions based on Lebesgue measure. Com- puters and Mathematics with Applications 56: 367-374 7. Pap E (1995) Null-additive set functions. Kluwer, Dordrecht 8. Ralescu D, Adams G (1980) The fuzzy integral. Journal of Mathematical Analysis and Applications 75: 562-570 9. Roman-Flores ´ H, Flores-Franulic ˇ A, Chalco-Cano Y (2007) The fuzzy integral for monotone func- tions. Applied Mathematics and Computation 185: 492-498 10. Roman-Flores ´ H, Flores-Franulic ˇ A, Chalco-Cano Y (2007) A Jensen type inequality for fuzzy inte- grals. Information Sciences 177: 3192-3201 11. Sugeno M (1974) Theory of fuzzy integrals and its applications. Ph.D thesis, Tokyo Institute of Technology 12. Wang Z, Klir G (1992) Fuzzy measure theory. Plenum Press, New York

Journal

Fuzzy Information and Engineering – Taylor & Francis

Published: Sep 1, 2010

Keywords: Sugeno integral; Fuzzy measure; Monotone function

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