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A Cauchy–Schwarz type inequality for fuzzy integrals
Nonlinear Analysis 73 (2010) 3329–3335
Contents lists available at ScienceDirect
Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
A Cauchy–Schwarz type inequality for fuzzy integrals J. Caballero ∗ , K. Sadarangani Departamento de Matemáticas, Universidad de Las Palmas de Gran Canaria, Campus de Tafira Baja, 35017 Las Palmas de Gran Canaria, Spain
article
info
Article history: Received 24 September 2009 Accepted 7 July 2010
abstract In this paper we prove a Cauchy–Schwarz type inequality for fuzzy integrals. © 2010 Elsevier Ltd. All rights reserved.
Keywords: Cauchy–Schwarz inequality Fuzzy integral Comonotone functions
1. Introduction The theory of fuzzy measures and fuzzy integrals was introduced by Sugeno [1] as a tool for modelling non-deterministic problems. The properties and applications of the Sugeno integral have been studied by many authors, among others, Ralescu and Adams [2] in the study of several equivalent definitions of fuzzy integrals, Román-Flores et al. [3,4] in studying level continuity of fuzzy integrals and H-continuity of fuzzy measures, and Wang and Klir [5] in a general overview on fuzzy measurement and fuzzy integration theory. Recently, some authors [6–16] have studied some fuzzy integral inequalities. The purpose of this paper is to prove a Cauchy–Schwarz type inequality for the Sugeno integral. Firstly, we introduce some basic notation and properties of the fuzzy integral. Suppose that Σ is a σ -algebra of subsets of X and let µ : Σ → [0, ∞] be a non-negative, extended real-valued set function. We say that µ is a fuzzy measure if it satisfies: (1) µ(∅) = 0. (2) E , F ∈ Σ and E ⊂ F imply µ(E ) ≤ µ(F ) (monotonicity). (3) {En } ⊂ Σ , E1 ⊂ E2 ⊂ · · ·, imply limn→∞ µ(En ) = µ ∪∞ n=1 En (continuity from below).
(4) {En } ⊂ Σ , E1 ⊃ E2 ⊃ · · ·, µ(E1 ) < ∞, imply limn→∞ µ(En ) = µ ∩∞ n=1 En (continuity from above).
The triple (X , Σ , µ) is called a fuzzy measure space. If f is a non-negative real-valued function defined on X , we will denote by Lα f = {x ∈ X : f (x) ≥ α} = {f ≥ α} the α -level of f , for α > 0, and L0 f = {x ∈ X : f (x) > 0} = suppf is the support of f . Note that if α ≤ β then {f ≥ β} ⊂ {f ≥ α}. Let (X , Σ , µ) be a fuzzy measure space; by F+ (X ), we denote the set of all non-negative measurable functions with respect to Σ . Definition 1 ([1]). Let (X , Σ , µ) be a fuzzy measure space, f ∈ F+ (X ) and A ∈ Σ ; the fuzzy integral (or Sugeno integral) of f on A with respect to the fuzzy measure µ is defined as
∗
Corresponding author. Tel.: +34 928 458 810; fax: +34 928 458 811. E-mail addresses: [email protected] (J. Caballero), [email protected] (K. Sadarangani).
0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.07.013
3330
J. Caballero, K. Sadarangani / Nonlinear Analysis 73 (2010) 3329–3335
f dµ = A
_
[α ∧ µ(A ∩ {f ≥ α})],
α≥0
where ∨ and ∧ denote the operations sup and inf on [0, ∞), respectively. It is well known that the Sugeno integral is a type of nonlinear integral, i.e., for general cases,
(af + bg )dµ = a f dµ + b gdµ does not hold. The following properties of the fuzzy integral are well known and can be found in [5]. Proposition 1. Let (X , Σ , µ) be a fuzzy measure space, A, B ∈ Σ and f , g ∈ F µ (X ); then: (1) (2) (3) (4) (5) (6) (7) (8)
f dµ ≤ µ(A). kdµ = k ∧ µ(A), k a non-negative constant. A If f ≤ g on A then A f dµ ≤ A gdµ. If A ⊂ B then A f dµ ≤ B f dµ. µ(A ∩ {f ≥ α}) ≥ α ⇒ A f dµ ≥ α . µ(A ∩ {f ≥ α}) ≤ α ⇒ A f dµ ≤ α . f dµ < α ⇔ there exists γ < α such that µ(A ∩ {f ≥ γ }) < α . A f dµ > α ⇔ there exists γ > α such that µ(A ∩ {f ≥ γ }) > α . A A
Remark 1. Let F be the function defined by F (α) = µ(A ∩ {f ≥ α}); then, due to (5) and (6) of Proposition 1, F (α) = α ⇒
f dµ = α. A
Thus, from a numerical point of view, the Sugeno integral can be calculated solving the equation F (α) = α . Notice that if the equation F (α) = α has a solution then the value of the Sugeno integral coincides with this solution. On the other hand, the existence of the Sugeno integral of a function f does not imply the existence of a solution of F (α) = α , as is shown in the following example. Example 1. Consider the function f : [0, ∞) −→ [0, ∞) given by
(
1 for 0 ≤ x ≤ 2 4 for x > 2 x2
f (x) =
and the classical Lebesgue measure µ in [0, ∞].
R∞
By definition, −0 f dµ =
W
α≥0 [α
∧ µ([0, ∞) ∩ {f ≥ α})]. Now, we calculate µ([0, ∞) ∩ {f ≥ α}) for every α ≥ 0.
Case 1. α > 1. In this case, as f (x) ≤ 1 for any x ∈ [0, ∞), then
µ([0, ∞) ∩ {f ≥ α}) = µ(∅) = 0. Case 2. α = 1. In this case, we have
µ([0, ∞) ∩ {f ≥ 1}) = µ([0, ∞) ∩ [0, 2]) = 2. Case 3. α < 1. In this case,
µ([0, ∞) ∩ {f ≥ α}) = µ([0, ∞) ∩ ({f ≥ 1} ∪ {x : α ≤ f (x) < 1})) 2 = µ [0, 2] ∪ [0, ∞) ∩ x > 2 : x ≤ √ α √ 2 = µ([0, 2/ α]) = √ . α Consequently, 0 if α > 1 (α ∧ µ([0, ∞) ∩ {f ≥ α})) = 1 if α = 1 α if α < 1 R∞ and this gives us that −0 f dµ = 1.
(
J. Caballero, K. Sadarangani / Nonlinear Analysis 73 (2010) 3329–3335
3331
On the other hand, taking into account the previous calculations we have
0 if α > 1 2 if α = 1 F (α) = √2 if α < 1 α and F (α) = α has no solution. 2. A Cauchy–Schwarz type inequality for fuzzy integrals The Cauchy–Schwarz inequality states that if (Ω , µ) is a measure space then
Z Ω
|f · g |dµ ≤
Z Ω
|f |2 dµ
21 Z 21 · |g |2 dµ , Ω
for f , g ∈ L2 (µ), where L2 (µ) = {f : Ω → R, measurable and Ω |f |2 dµ < ∞}. Unfortunately, the following example proves that, in general, the Cauchy–Schwarz inequality is not valid in the fuzzy context.
R
Example 2. Consider X = [0, 1] and let µ be the usual Lebesgue measure on X . Note that for any f ∈ F+ (X ), by (1) of Proposition 1, X f 2 dµ ≤ µ(X ) = 1. If we take the function f defined by
( f ( x) =
0, 1 x
x=0
,
x ∈ (0, 1]
and the function g (x) = x, then a straightforward calculus shows that 0.3819 and
1
g 2 (x)dx
12 ·
0
1
f 2 (x)dx
12
1
< 0.618 < 1 =
0
1 0
f (x) · g (x)dx = 1,
1 0
1 0
f 2 (x)dx = 1,
g 2 (x)dx ≈
f (x)g (x)dx.
0
In order to prove a Cauchy–Schwarz type inequality for fuzzy integrals we need the following lemma. Lemma 1. Let (X , Σ , µ) be a fuzzy measure space, f ∈ F+ (X ), A ∈ Σ , f s dµ ≥
A
f dµ
s
A
f dµ ≤ 1 and s ≥ 1. Then
.
A
Proof. Put a = A f dµ. If a = 0 then the result is trivial. Suppose a > 0. Let α be such that a > α > 0. Then, by (8) of Proposition 1, there exists γ > α such that µ(A ∩{f ≥ γ }) > α . As {f ≥ γ } = {f s ≥ γ s }, s ≥ 1 and, by (1) of Proposition 1, 0 < α < a = A f dµ ≤ 1, we can obtain
µ(A ∩ {f s ≥ γ s }) = µ(A ∩ {f ≥ γ }) > α > α s . As γ s > α s , applying (8) of Proposition 1, we have f s dµ > α s . A
Since the value α is such that A f dµ = a > α > 0 and arbitrary, in particular, if we take αn = n ≥ n0 for a certain n0 ∈ N, we get f s dµ > A
f dµ − A
1 n
s
A
f dµ −
1 n
> 0, where
for n ≥ n0 .
Finally, taking the limit when n → ∞, we obtain the result.
Remark 2. Notice that Lemma 1 is a particular case of the Jensen inequality which appears in Theorem 1 of [13]. The following example shows that the condition
A
f dµ ≤ 1 is necessary in Lemma 1.
Example 3. Consider X = [0, 2] and let µ be the usual Lebesgue measure on X . We take f (x) = 4 for x ∈ X and s = 2; then, by (2) of Proposition 1,
3332
J. Caballero, K. Sadarangani / Nonlinear Analysis 73 (2010) 3329–3335 2
f 2 dµ = and (
X
16 dx = 2
X
0
f dµ)2 = (
2 4dx 2 0
) = 22 = 4.
Now, we present the following definition. Definition 2. Two functions f , g : X → R are said to be comonotone if for all (x, y) ∈ X 2 ,
(f (x) − f (y))(g (x) − g (y)) ≥ 0. An important property of comonotone functions is that for any real numbers p, q, either {f ≥ p} ⊂ {g ≥ q} or {g ≥ q} ⊂ {f ≥ p}. Note that two monotone functions (in the same sense) are comonotone. Lemma 2. Let f , g : X → [0, ∞) be two comonotone functions. Then the functions f 2 and g 2 are comonotone. Proof. For x, y ∈ X we have
(f 2 (x) − f 2 (y))(g 2 (x) − g 2 (y)) = (f (x) − f (y))(g (x) − g (y)) · (f (x) + f (y))(g (x) + g (y)). From this and as f and g are non-negative and comonotone, the lemma is obvious.
Now, we will show a Cauchy–Schwarz type inequality for the Sugeno integral. Theorem 1. Let (X , Σ , µ) be a fuzzy measure space, f , g ∈ F+ (X ) and f and g comonotone functions, A ∈ Σ with f · g dµ ≤ 1. Then
f · g dµ
2
f 2 dµ ∨
≤
A
A
A
g 2 dµ . A
Proof. By Lemma 1,
Put a =
f · g dµ
2
f 2 · g 2 dµ.
≤
A
A
A
f 2 g 2 dµ. We consider two cases.
(1)
Case 1. a > 0. If A f · gdµ < 1, by (7) of Proposition 1, there exists γ < 1 such that µ(A ∩ {f · g ≥ γ }) < 1. Then, as µ(A ∩ {f · g ≥ γ }) = µ({A ∩ {f 2 · g 2 ≥ γ 2 }}) < 1 and γ < 1, applying again property (7) we conclude that A f 2 · g 2 dµ < 1. Let α be such that 0 < α < a ≤ 1. Then by (8) of Proposition 1, there exists γα > α (note that we can choose γα ≤ 1) such that
µ(A ∩ {f 2 · g 2 ≥ γα }) > α. (2) √ √ As {f 2 · g 2 ≥ γα } ⊂ {f 2 ≥ γα } ∪ {g 2 ≥ γα } and, by Lemma 2, f 2 and g 2 are comonotone functions, then either √ √ √ √ √ √ 2 2 2 {f ≥ γα } ⊂ {g ≥ γα } or {g ≥ γα } ⊂ {f 2 ≥ γα }. Suppose that {f 2 ≥ γα } ⊂ {g 2 ≥ γα }. In this case {f 2 g 2 ≥ γα } ⊂ {g 2 ≥
√ γα }.
From (2) we have
√ µ A ∩ g 2 ≥ γα ≥ µ(A ∩ {f 2 · g 2 ≥ γα }) > α. √ As γα ≤ 1, α < γα ≤ γα < 1 and, using (8) of Proposition 1, we obtain g s dµ > α. A
As 0 < α < a and arbitrary, we have f 2 g 2 dµ ≤
a= A
g 2 dµ. A
For the case {g 2 ≥
√ √ γα } ⊂ {f 2 ≥ γα } we use a similar argument and we can prove
f 2 g 2 dµ ≤
a= A
f 2 dµ. A
J. Caballero, K. Sadarangani / Nonlinear Analysis 73 (2010) 3329–3335
3333
This gives us
f g dµ ≤ 2 2
a=
f dµ ∨
A
g dµ .
2
2
A
A
Taking into account (1) and the last inequality we get the desired result. Case 2. a = 0. In this case, A f 2 g 2 dµ = 0. Using Theorem 6 in [11], we have a.e. on A,
f 2 · g2 = 0 or, equivalently, f ·g =0
a.e. on A.
Again, by Theorem 6 in [11] f · g dµ = 0 A
and, consequently, the desired result is obvious.
Remark 3. The inequality appearing in Theorem 1 is sharp. In fact, if we consider X = [0, 1], µ the usual Lebesgue measure and f = g = 1 on X , then it is easily seen that 1
f (x)g (x)dx
2
1
=
0
1
f 2 (x)dx ∨
0
g 2 (x)dx .
0
In the sequel we will prove a Cauchy–Schwarz type inequality for the case f , g : X → [0, 1]. Theorem 2. Let (X , Σ , µ) be a fuzzy measure space, f , g : X → [0, 1] measurable functions and A ∈ Σ . Then
f · g dµ
2
f 2 dµ
≤
A
g 2 dµ .
∧
A
A
Proof. As f , g : X → [0, 1], by properties (3) and (2) of Proposition 1 we have f · gdµ ≤
1 dµ ≤ 1.
A
A
Now, by Lemma 1, we obtain
f · gdµ
2
f 2 · g 2 dµ.
≤
A
(3)
A
Taking into account that f 2 , g 2 : X → [0, 1], we have f 2 · g2 ≤ f 2 f 2 · g 2 ≤ g 2. By (3) of Proposition 1, the last inequalities imply f 2 · g 2 dµ ≤ A
f 2 dµ A
f 2 · g 2 dµ ≤ A
Hence,
A
g 2 dµ. A
f 2 · g 2 dµ ≤ (
A
f 2 dµ) ∧ (
g 2 dµ). This inequality and (3) give us the desired result.
A
3. Applications Now, we present some applications of our results. Firstly, we recall the following result which appears in Theorems 2.1 and 2.2 of [10]. Theorem 3 ([10] (Chebyshev’s Inequality)). Suppose that f and g are two real-valued functions from [0, 1] to [0, ∞) and that µ is the Lebesgue measure on R. If f and g are increasing (or decreasing) functions then 1 0
f · gdµ ≥
1 0
f dµ ·
1 0
gdµ .
3334
J. Caballero, K. Sadarangani / Nonlinear Analysis 73 (2010) 3329–3335
Taking into account Theorems 1 and 3 and the fact that two functions monotone in the same sense are comonotone, we have the following result. Theorem 4. Suppose that f and g are two real-valued functions from [0, 1] to [0, ∞) and that µ is the Lebesgue measure on R. If f and g are monotone in the same sense (both increasing or decreasing) then 1
f dµ ·
0
1
1
gdµ ≤
0
s f · gdµ ≤
0
1
1
f 2 dµ ∨
0
g 2 dµ.
0
Theorem 5. Let f , g : [0, ∞) → [0, ∞) be measurable functions, µ the Lebesgue measure on R and x 0
If F (x) =
∞
x 0
f (t ) dµ and G(x) = F (x) · G(x) dµ
g (t ) dµ then
2
∞
f dµ ≤ 1,
∞ 0
g dµ < 1.
∞
f 2 (x) dµ ∧
≤
0
∞ 0
g 2 (x) dµ.
0
0
Proof. Obviously, as f and g are non-negative functions, by (4) of Proposition 1, F (x) and G(x) are increasing functions and, consequently, F (x) and G(x) are comonotone functions. ∞ ∞ Moreover, by our assumption 0 f dµ ≤ 1 and 0 g dµ ≤ 1, we have that F (x) ≤ 1 and G(x) ≤ 1. By Theorem 2 ∞
F (x) · G(x) dµ
2
∞
0
As
∞ 0
∞
F 2 ( x ) dµ ∧
≤ 0
G2 (x) dµ.
(4)
0
f dµ ≤ 1, by Lemma 2 of [16] and by (4) of Proposition 1, we have ∞
f (x) dµ ≥ 2
∞
f ( x ) dµ
0
2
x
≥
0
Therefore, if q =
f (t ) dµ
2
= F 2 (x).
0
∞ 2 0
f (x) dµ, then
∞
∞
q dµ ≥
F 2 (x) dµ.
0
0
By (2) of Proposition 1, we get ∞
∞
f 2 (x) dµ = q =
∞
q dµ ≥
0
0
F 2 (x) dµ.
(5)
0
Similarly, we can get ∞
∞
G2 (x) dµ ≤ 0
g 2 (x) dµ.
(6)
0
From (4), (5) and (6) we have ∞
F (x) · G(x) dµ
2
∞
∞
f 2 (x) dµ ∧
≤
0
0
g 2 (x) dµ. 0
∞
∞
If in Theorem 5 we drop the condition 0 f dµ ≤ 1 and 0 g dµ ≤ 1 we have the following result. Previously, we recalled the following fuzzy Hardy inequality which appears in Theorem 2 in [16].
Rx
Theorem 6 (Fuzzy Hardy Inequality: Infinite Case). Let f : [0, ∞) −→ [0, ∞) be an integrable function and F (x) = −0 f (t )dt; then
Z ∞ p p+1 1 Z ∞ F (x) − f (x)p dx ≥− dx, x
0
0
for all p ≥ 1. Theorem 7. Let f , g : [0, ∞) → [0, ∞) be measurable functions and µ the Lebesgue measure on R. If F (x) = G(x) =
x 0
g (t ) dµ then ∞
0
F (x) x
·
G(x) x
dµ
2
∞
f 2 dµ
≤ 0
31
∞
g 2 dµ
∧ 0
13
.
x 0
f (t ) dµ and
J. Caballero, K. Sadarangani / Nonlinear Analysis 73 (2010) 3329–3335 x 0
Proof. Firstly, note that due to (1) of Proposition 1, F (x) = F (x) x
≤ 1 and
G(x) x
x
0
x 0
g (t )dµ ≤ x, and consequently,
≤ 1. By Theorem 2, we have
F (x)
∞
f (t )dµ ≤ x and G(x) =
3335
·
G(x) x
dµ
2
∞
≤
F (x)
2
x
0
∞
dµ ∧
G(x) x
0
2
dµ.
(7)
Taking this into Theorem 6, ∞
F (x)
2
x
0
∞
G(x)
f (x) dµ 2
13
0
2
x
0
dµ ≤
∞
dµ ≤
∞
g 2 (x) dµ
31
.
0
Therefore, the last inequalities and (7) give us ∞
0
F (x) x
·
G(x) x
dµ
2
∞
f dµ 2
≤ 0
13
∞
g dµ 2
∧
13
.
0
Acknowledgement This research was partially supported by the ‘‘Ministerio de Educación y Ciencia’’, Project MTM 2007/65706. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]
M. Sugeno, Theory of fuzzy integrals and its applications, Ph.D. Thesis, Tokyo Institute of Technology, 1974. D. Ralescu, G. Adams, The fuzzy integral, J. Math. Anal. Appl. 75 (1980) 562–570. H. Román-Flores, A. Flores-Franuli˘c, R. Bassanezi, M. Rojas-Medar, On the level-continuity of fuzzy integrals, Fuzzy Sets Syst. 80 (1996) 339–344. H. Román-Flores, Y. Chalco-Cano, H-Continuity of fuzzy measures and set defuzzification, Fuzzy Sets Syst. 157 (2006) 230–242. Z. Wang, G. Klir, Fuzzy Measure Theory, Plenum, New York, 1992. H. Agahi, R. Mesiar, Y. Ouyang, General Minkowski type inequalities for Sugeno integrals, Fuzzy Sets Syst. 161 (2010) 708–715. H. Agahi, R. Mesiar, Y. Ouyang, New general extensions of Chebyshev type inequalities for Sugeno integrals, Int. J. Approx. Reason. 51 (2009) 135–140. A. Flores-Franuli˘c, H. Román-Flores, A Chebyshev type inequality for fuzzy integrals, Appl. Math. Comput. 190 (2007) 1178–1184. A. Flores-Franuli˘c, H. Román-Flores, Y. Chalco-Cano, Markov type inequalities for fuzzy integrals, Appl. Math. Comput. 207 (2009) 242–247. Q.S. Mao, A note on the Chebyshev-type inequality of Sugeno integrals, Appl. Math. Comput. 212 (2009) 275–279. R. Mesiar, Y. Ouyang, General Chebyshev type inequalities for Sugeno integrals, Fuzzy Sets Syst. 160 (2009) 58–64. Y. Ouyang, J. Fang, Sugeno integral of monotone functions based on Lebesgue measure, Comput. Math. Appl. 56 (2008) 367–374. H. Román-Flores, A. Flores-Franuli˘c, Y. Chalco-Cano, A Jensen type inequality for fuzzy integrals, Inform. Sci. 177 (15) (2007) 3192–3201. H. Román-Flores, A. Flores-Franuli˘c, Y. Chalco-Cano, The fuzzy integral for monotone functions, Appl. Math. Comput. 185 (2007) 492–498. H. Román-Flores, A. Flores-Franuli˘c, Y. Chalco-Cano, A convolution type inequality for fuzzy integrals, Appl. Math. Comput. 195 (2008) 94–99. H. Román-Flores, A. Flores-Franuli˘c, Y. Chalco-Cano, A Hardy-type inequality for fuzzy integrals, Appl. Math. Comput. 204 (2008) 178–183.
Contents lists available at ScienceDirect
Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
A Cauchy–Schwarz type inequality for fuzzy integrals J. Caballero ∗ , K. Sadarangani Departamento de Matemáticas, Universidad de Las Palmas de Gran Canaria, Campus de Tafira Baja, 35017 Las Palmas de Gran Canaria, Spain
article
info
Article history: Received 24 September 2009 Accepted 7 July 2010
abstract In this paper we prove a Cauchy–Schwarz type inequality for fuzzy integrals. © 2010 Elsevier Ltd. All rights reserved.
Keywords: Cauchy–Schwarz inequality Fuzzy integral Comonotone functions
1. Introduction The theory of fuzzy measures and fuzzy integrals was introduced by Sugeno [1] as a tool for modelling non-deterministic problems. The properties and applications of the Sugeno integral have been studied by many authors, among others, Ralescu and Adams [2] in the study of several equivalent definitions of fuzzy integrals, Román-Flores et al. [3,4] in studying level continuity of fuzzy integrals and H-continuity of fuzzy measures, and Wang and Klir [5] in a general overview on fuzzy measurement and fuzzy integration theory. Recently, some authors [6–16] have studied some fuzzy integral inequalities. The purpose of this paper is to prove a Cauchy–Schwarz type inequality for the Sugeno integral. Firstly, we introduce some basic notation and properties of the fuzzy integral. Suppose that Σ is a σ -algebra of subsets of X and let µ : Σ → [0, ∞] be a non-negative, extended real-valued set function. We say that µ is a fuzzy measure if it satisfies: (1) µ(∅) = 0. (2) E , F ∈ Σ and E ⊂ F imply µ(E ) ≤ µ(F ) (monotonicity). (3) {En } ⊂ Σ , E1 ⊂ E2 ⊂ · · ·, imply limn→∞ µ(En ) = µ ∪∞ n=1 En (continuity from below).
(4) {En } ⊂ Σ , E1 ⊃ E2 ⊃ · · ·, µ(E1 ) < ∞, imply limn→∞ µ(En ) = µ ∩∞ n=1 En (continuity from above).
The triple (X , Σ , µ) is called a fuzzy measure space. If f is a non-negative real-valued function defined on X , we will denote by Lα f = {x ∈ X : f (x) ≥ α} = {f ≥ α} the α -level of f , for α > 0, and L0 f = {x ∈ X : f (x) > 0} = suppf is the support of f . Note that if α ≤ β then {f ≥ β} ⊂ {f ≥ α}. Let (X , Σ , µ) be a fuzzy measure space; by F+ (X ), we denote the set of all non-negative measurable functions with respect to Σ . Definition 1 ([1]). Let (X , Σ , µ) be a fuzzy measure space, f ∈ F+ (X ) and A ∈ Σ ; the fuzzy integral (or Sugeno integral) of f on A with respect to the fuzzy measure µ is defined as
∗
Corresponding author. Tel.: +34 928 458 810; fax: +34 928 458 811. E-mail addresses: [email protected] (J. Caballero), [email protected] (K. Sadarangani).
0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.07.013
3330
J. Caballero, K. Sadarangani / Nonlinear Analysis 73 (2010) 3329–3335
f dµ = A
_
[α ∧ µ(A ∩ {f ≥ α})],
α≥0
where ∨ and ∧ denote the operations sup and inf on [0, ∞), respectively. It is well known that the Sugeno integral is a type of nonlinear integral, i.e., for general cases,
(af + bg )dµ = a f dµ + b gdµ does not hold. The following properties of the fuzzy integral are well known and can be found in [5]. Proposition 1. Let (X , Σ , µ) be a fuzzy measure space, A, B ∈ Σ and f , g ∈ F µ (X ); then: (1) (2) (3) (4) (5) (6) (7) (8)
f dµ ≤ µ(A). kdµ = k ∧ µ(A), k a non-negative constant. A If f ≤ g on A then A f dµ ≤ A gdµ. If A ⊂ B then A f dµ ≤ B f dµ. µ(A ∩ {f ≥ α}) ≥ α ⇒ A f dµ ≥ α . µ(A ∩ {f ≥ α}) ≤ α ⇒ A f dµ ≤ α . f dµ < α ⇔ there exists γ < α such that µ(A ∩ {f ≥ γ }) < α . A f dµ > α ⇔ there exists γ > α such that µ(A ∩ {f ≥ γ }) > α . A A
Remark 1. Let F be the function defined by F (α) = µ(A ∩ {f ≥ α}); then, due to (5) and (6) of Proposition 1, F (α) = α ⇒
f dµ = α. A
Thus, from a numerical point of view, the Sugeno integral can be calculated solving the equation F (α) = α . Notice that if the equation F (α) = α has a solution then the value of the Sugeno integral coincides with this solution. On the other hand, the existence of the Sugeno integral of a function f does not imply the existence of a solution of F (α) = α , as is shown in the following example. Example 1. Consider the function f : [0, ∞) −→ [0, ∞) given by
(
1 for 0 ≤ x ≤ 2 4 for x > 2 x2
f (x) =
and the classical Lebesgue measure µ in [0, ∞].
R∞
By definition, −0 f dµ =
W
α≥0 [α
∧ µ([0, ∞) ∩ {f ≥ α})]. Now, we calculate µ([0, ∞) ∩ {f ≥ α}) for every α ≥ 0.
Case 1. α > 1. In this case, as f (x) ≤ 1 for any x ∈ [0, ∞), then
µ([0, ∞) ∩ {f ≥ α}) = µ(∅) = 0. Case 2. α = 1. In this case, we have
µ([0, ∞) ∩ {f ≥ 1}) = µ([0, ∞) ∩ [0, 2]) = 2. Case 3. α < 1. In this case,
µ([0, ∞) ∩ {f ≥ α}) = µ([0, ∞) ∩ ({f ≥ 1} ∪ {x : α ≤ f (x) < 1})) 2 = µ [0, 2] ∪ [0, ∞) ∩ x > 2 : x ≤ √ α √ 2 = µ([0, 2/ α]) = √ . α Consequently, 0 if α > 1 (α ∧ µ([0, ∞) ∩ {f ≥ α})) = 1 if α = 1 α if α < 1 R∞ and this gives us that −0 f dµ = 1.
(
J. Caballero, K. Sadarangani / Nonlinear Analysis 73 (2010) 3329–3335
3331
On the other hand, taking into account the previous calculations we have
0 if α > 1 2 if α = 1 F (α) = √2 if α < 1 α and F (α) = α has no solution. 2. A Cauchy–Schwarz type inequality for fuzzy integrals The Cauchy–Schwarz inequality states that if (Ω , µ) is a measure space then
Z Ω
|f · g |dµ ≤
Z Ω
|f |2 dµ
21 Z 21 · |g |2 dµ , Ω
for f , g ∈ L2 (µ), where L2 (µ) = {f : Ω → R, measurable and Ω |f |2 dµ < ∞}. Unfortunately, the following example proves that, in general, the Cauchy–Schwarz inequality is not valid in the fuzzy context.
R
Example 2. Consider X = [0, 1] and let µ be the usual Lebesgue measure on X . Note that for any f ∈ F+ (X ), by (1) of Proposition 1, X f 2 dµ ≤ µ(X ) = 1. If we take the function f defined by
( f ( x) =
0, 1 x
x=0
,
x ∈ (0, 1]
and the function g (x) = x, then a straightforward calculus shows that 0.3819 and
1
g 2 (x)dx
12 ·
0
1
f 2 (x)dx
12
1
< 0.618 < 1 =
0
1 0
f (x) · g (x)dx = 1,
1 0
1 0
f 2 (x)dx = 1,
g 2 (x)dx ≈
f (x)g (x)dx.
0
In order to prove a Cauchy–Schwarz type inequality for fuzzy integrals we need the following lemma. Lemma 1. Let (X , Σ , µ) be a fuzzy measure space, f ∈ F+ (X ), A ∈ Σ , f s dµ ≥
A
f dµ
s
A
f dµ ≤ 1 and s ≥ 1. Then
.
A
Proof. Put a = A f dµ. If a = 0 then the result is trivial. Suppose a > 0. Let α be such that a > α > 0. Then, by (8) of Proposition 1, there exists γ > α such that µ(A ∩{f ≥ γ }) > α . As {f ≥ γ } = {f s ≥ γ s }, s ≥ 1 and, by (1) of Proposition 1, 0 < α < a = A f dµ ≤ 1, we can obtain
µ(A ∩ {f s ≥ γ s }) = µ(A ∩ {f ≥ γ }) > α > α s . As γ s > α s , applying (8) of Proposition 1, we have f s dµ > α s . A
Since the value α is such that A f dµ = a > α > 0 and arbitrary, in particular, if we take αn = n ≥ n0 for a certain n0 ∈ N, we get f s dµ > A
f dµ − A
1 n
s
A
f dµ −
1 n
> 0, where
for n ≥ n0 .
Finally, taking the limit when n → ∞, we obtain the result.
Remark 2. Notice that Lemma 1 is a particular case of the Jensen inequality which appears in Theorem 1 of [13]. The following example shows that the condition
A
f dµ ≤ 1 is necessary in Lemma 1.
Example 3. Consider X = [0, 2] and let µ be the usual Lebesgue measure on X . We take f (x) = 4 for x ∈ X and s = 2; then, by (2) of Proposition 1,
3332
J. Caballero, K. Sadarangani / Nonlinear Analysis 73 (2010) 3329–3335 2
f 2 dµ = and (
X
16 dx = 2
X
0
f dµ)2 = (
2 4dx 2 0
) = 22 = 4.
Now, we present the following definition. Definition 2. Two functions f , g : X → R are said to be comonotone if for all (x, y) ∈ X 2 ,
(f (x) − f (y))(g (x) − g (y)) ≥ 0. An important property of comonotone functions is that for any real numbers p, q, either {f ≥ p} ⊂ {g ≥ q} or {g ≥ q} ⊂ {f ≥ p}. Note that two monotone functions (in the same sense) are comonotone. Lemma 2. Let f , g : X → [0, ∞) be two comonotone functions. Then the functions f 2 and g 2 are comonotone. Proof. For x, y ∈ X we have
(f 2 (x) − f 2 (y))(g 2 (x) − g 2 (y)) = (f (x) − f (y))(g (x) − g (y)) · (f (x) + f (y))(g (x) + g (y)). From this and as f and g are non-negative and comonotone, the lemma is obvious.
Now, we will show a Cauchy–Schwarz type inequality for the Sugeno integral. Theorem 1. Let (X , Σ , µ) be a fuzzy measure space, f , g ∈ F+ (X ) and f and g comonotone functions, A ∈ Σ with f · g dµ ≤ 1. Then
f · g dµ
2
f 2 dµ ∨
≤
A
A
A
g 2 dµ . A
Proof. By Lemma 1,
Put a =
f · g dµ
2
f 2 · g 2 dµ.
≤
A
A
A
f 2 g 2 dµ. We consider two cases.
(1)
Case 1. a > 0. If A f · gdµ < 1, by (7) of Proposition 1, there exists γ < 1 such that µ(A ∩ {f · g ≥ γ }) < 1. Then, as µ(A ∩ {f · g ≥ γ }) = µ({A ∩ {f 2 · g 2 ≥ γ 2 }}) < 1 and γ < 1, applying again property (7) we conclude that A f 2 · g 2 dµ < 1. Let α be such that 0 < α < a ≤ 1. Then by (8) of Proposition 1, there exists γα > α (note that we can choose γα ≤ 1) such that
µ(A ∩ {f 2 · g 2 ≥ γα }) > α. (2) √ √ As {f 2 · g 2 ≥ γα } ⊂ {f 2 ≥ γα } ∪ {g 2 ≥ γα } and, by Lemma 2, f 2 and g 2 are comonotone functions, then either √ √ √ √ √ √ 2 2 2 {f ≥ γα } ⊂ {g ≥ γα } or {g ≥ γα } ⊂ {f 2 ≥ γα }. Suppose that {f 2 ≥ γα } ⊂ {g 2 ≥ γα }. In this case {f 2 g 2 ≥ γα } ⊂ {g 2 ≥
√ γα }.
From (2) we have
√ µ A ∩ g 2 ≥ γα ≥ µ(A ∩ {f 2 · g 2 ≥ γα }) > α. √ As γα ≤ 1, α < γα ≤ γα < 1 and, using (8) of Proposition 1, we obtain g s dµ > α. A
As 0 < α < a and arbitrary, we have f 2 g 2 dµ ≤
a= A
g 2 dµ. A
For the case {g 2 ≥
√ √ γα } ⊂ {f 2 ≥ γα } we use a similar argument and we can prove
f 2 g 2 dµ ≤
a= A
f 2 dµ. A
J. Caballero, K. Sadarangani / Nonlinear Analysis 73 (2010) 3329–3335
3333
This gives us
f g dµ ≤ 2 2
a=
f dµ ∨
A
g dµ .
2
2
A
A
Taking into account (1) and the last inequality we get the desired result. Case 2. a = 0. In this case, A f 2 g 2 dµ = 0. Using Theorem 6 in [11], we have a.e. on A,
f 2 · g2 = 0 or, equivalently, f ·g =0
a.e. on A.
Again, by Theorem 6 in [11] f · g dµ = 0 A
and, consequently, the desired result is obvious.
Remark 3. The inequality appearing in Theorem 1 is sharp. In fact, if we consider X = [0, 1], µ the usual Lebesgue measure and f = g = 1 on X , then it is easily seen that 1
f (x)g (x)dx
2
1
=
0
1
f 2 (x)dx ∨
0
g 2 (x)dx .
0
In the sequel we will prove a Cauchy–Schwarz type inequality for the case f , g : X → [0, 1]. Theorem 2. Let (X , Σ , µ) be a fuzzy measure space, f , g : X → [0, 1] measurable functions and A ∈ Σ . Then
f · g dµ
2
f 2 dµ
≤
A
g 2 dµ .
∧
A
A
Proof. As f , g : X → [0, 1], by properties (3) and (2) of Proposition 1 we have f · gdµ ≤
1 dµ ≤ 1.
A
A
Now, by Lemma 1, we obtain
f · gdµ
2
f 2 · g 2 dµ.
≤
A
(3)
A
Taking into account that f 2 , g 2 : X → [0, 1], we have f 2 · g2 ≤ f 2 f 2 · g 2 ≤ g 2. By (3) of Proposition 1, the last inequalities imply f 2 · g 2 dµ ≤ A
f 2 dµ A
f 2 · g 2 dµ ≤ A
Hence,
A
g 2 dµ. A
f 2 · g 2 dµ ≤ (
A
f 2 dµ) ∧ (
g 2 dµ). This inequality and (3) give us the desired result.
A
3. Applications Now, we present some applications of our results. Firstly, we recall the following result which appears in Theorems 2.1 and 2.2 of [10]. Theorem 3 ([10] (Chebyshev’s Inequality)). Suppose that f and g are two real-valued functions from [0, 1] to [0, ∞) and that µ is the Lebesgue measure on R. If f and g are increasing (or decreasing) functions then 1 0
f · gdµ ≥
1 0
f dµ ·
1 0
gdµ .
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J. Caballero, K. Sadarangani / Nonlinear Analysis 73 (2010) 3329–3335
Taking into account Theorems 1 and 3 and the fact that two functions monotone in the same sense are comonotone, we have the following result. Theorem 4. Suppose that f and g are two real-valued functions from [0, 1] to [0, ∞) and that µ is the Lebesgue measure on R. If f and g are monotone in the same sense (both increasing or decreasing) then 1
f dµ ·
0
1
1
gdµ ≤
0
s f · gdµ ≤
0
1
1
f 2 dµ ∨
0
g 2 dµ.
0
Theorem 5. Let f , g : [0, ∞) → [0, ∞) be measurable functions, µ the Lebesgue measure on R and x 0
If F (x) =
∞
x 0
f (t ) dµ and G(x) = F (x) · G(x) dµ
g (t ) dµ then
2
∞
f dµ ≤ 1,
∞ 0
g dµ < 1.
∞
f 2 (x) dµ ∧
≤
0
∞ 0
g 2 (x) dµ.
0
0
Proof. Obviously, as f and g are non-negative functions, by (4) of Proposition 1, F (x) and G(x) are increasing functions and, consequently, F (x) and G(x) are comonotone functions. ∞ ∞ Moreover, by our assumption 0 f dµ ≤ 1 and 0 g dµ ≤ 1, we have that F (x) ≤ 1 and G(x) ≤ 1. By Theorem 2 ∞
F (x) · G(x) dµ
2
∞
0
As
∞ 0
∞
F 2 ( x ) dµ ∧
≤ 0
G2 (x) dµ.
(4)
0
f dµ ≤ 1, by Lemma 2 of [16] and by (4) of Proposition 1, we have ∞
f (x) dµ ≥ 2
∞
f ( x ) dµ
0
2
x
≥
0
Therefore, if q =
f (t ) dµ
2
= F 2 (x).
0
∞ 2 0
f (x) dµ, then
∞
∞
q dµ ≥
F 2 (x) dµ.
0
0
By (2) of Proposition 1, we get ∞
∞
f 2 (x) dµ = q =
∞
q dµ ≥
0
0
F 2 (x) dµ.
(5)
0
Similarly, we can get ∞
∞
G2 (x) dµ ≤ 0
g 2 (x) dµ.
(6)
0
From (4), (5) and (6) we have ∞
F (x) · G(x) dµ
2
∞
∞
f 2 (x) dµ ∧
≤
0
0
g 2 (x) dµ. 0
∞
∞
If in Theorem 5 we drop the condition 0 f dµ ≤ 1 and 0 g dµ ≤ 1 we have the following result. Previously, we recalled the following fuzzy Hardy inequality which appears in Theorem 2 in [16].
Rx
Theorem 6 (Fuzzy Hardy Inequality: Infinite Case). Let f : [0, ∞) −→ [0, ∞) be an integrable function and F (x) = −0 f (t )dt; then
Z ∞ p p+1 1 Z ∞ F (x) − f (x)p dx ≥− dx, x
0
0
for all p ≥ 1. Theorem 7. Let f , g : [0, ∞) → [0, ∞) be measurable functions and µ the Lebesgue measure on R. If F (x) = G(x) =
x 0
g (t ) dµ then ∞
0
F (x) x
·
G(x) x
dµ
2
∞
f 2 dµ
≤ 0
31
∞
g 2 dµ
∧ 0
13
.
x 0
f (t ) dµ and
J. Caballero, K. Sadarangani / Nonlinear Analysis 73 (2010) 3329–3335 x 0
Proof. Firstly, note that due to (1) of Proposition 1, F (x) = F (x) x
≤ 1 and
G(x) x
x
0
x 0
g (t )dµ ≤ x, and consequently,
≤ 1. By Theorem 2, we have
F (x)
∞
f (t )dµ ≤ x and G(x) =
3335
·
G(x) x
dµ
2
∞
≤
F (x)
2
x
0
∞
dµ ∧
G(x) x
0
2
dµ.
(7)
Taking this into Theorem 6, ∞
F (x)
2
x
0
∞
G(x)
f (x) dµ 2
13
0
2
x
0
dµ ≤
∞
dµ ≤
∞
g 2 (x) dµ
31
.
0
Therefore, the last inequalities and (7) give us ∞
0
F (x) x
·
G(x) x
dµ
2
∞
f dµ 2
≤ 0
13
∞
g dµ 2
∧
13
.
0
Acknowledgement This research was partially supported by the ‘‘Ministerio de Educación y Ciencia’’, Project MTM 2007/65706. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]
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