On Chebyshev type inequalities for generalized Sugeno integrals
On Chebyshev type inequalities for generalized Sugeno integrals Marek Kaluszka1 , Andrzej Okolewski, Michal Boczek Institute of Mathematics, Lodz University of Technology, 90-924 Lodz, Poland
Abstract We give the necessary and su cient conditions guaranteeing the validity of Chebyshev type inequalities for generalized Sugeno integrals in the case of functions belonging to a much wider class than the comonotone functions. For several choices of operators, we characterize the classes of functions for which the Chebyshev type inequality for the classical Sugeno integral is satis ed.
1
Introduction
The pioneering concept of the fuzzy integral was introduced by Sugeno [24] as a tool for modeling non-deterministic problems. Theoretical investigations of the integral and its generalizations have been pursued by many researchers. Among others, Ralescu and Adams [18] provided several characterizations of the Sugeno integral and proved a monotone convergence theorem for this integral, RománFlores et al. [19, 20] discussed level-continuity of fuzzy integrals and H-continuity of fuzzy measures, while Wang and Klir [26] presented an excellent general overview on fuzzy measurement and fuzzy integration theory. On the other hand, fuzzy integrals have also been successfully applied to various elds (see, e.g., [6], [8],
[15], [25]).
Corresponding author. E-mail adress: [email protected]; tel.: +48 42 6313859; fax.: +48 42 6363114. 1
1
The study of inequalities for Sugeno integral was initiated by Román-Flores et al. [21]. Since then, the fuzzy integral counterparts of several classical inequalities, including Chebyshev's, Jensen's, Minkowski's and Hölder's inequalities, are given by Flores-Franuli¢ and Román-Flores [7], Agahi et al. [2, 3, 4], Mesiar and Ouyang
[12], Román-Flores et al. [22], and other researchers. The purpose of this paper is to study not only the su cient conditions for the validity of Chebyshev type and reverse Chebyshev type inequalities for the generalized Sugeno integral (cf. [1], [3], [5],[7], [12] and [16]), but also the necessary ones. The results are obtained for quite universal integrals and a rich class of functions, including the comonotone functions as a special case. For some speci c choices of operators, the corresponding characterizations of classes of functions for which Chebyshev's inequality for the classical Sugeno integral is satis ed are also presented. The paper is organized as follows. In Sections 2 and 3 we set up notation and terminology and present our main results, some related results as well as several illustrative examples. Concluding remarks are given in Section 4.
2
Main results
Let (X, F) be a measurable space and µ : F → [0, ∞] be a monotone measure, i.e., µ(∅) = 0, µ(X) > 0 and µ(A) 6 µ(B) whenever A ⊂ B. In the literature monotone measures are also called fuzzy measures (see, e.g., [26]). Let Y ⊂ R be an arbitrary nonempty interval; usually Y = [0, 1], Y = [0, ∞], Y = [0, ∞) or
Y = R. We denote the range of µ by µ(F). For a measurable function h : X → Y, we will de ne the Sugeno integral of h on a set A ∈ F with respect to µ and an
2
operator O : Y × µ(F) → Y as Z h O µ = sup a O µ(A ∩ {h > a}) ,
(1)
a∈Y A
where {h > a} stands for Z {x ∈ X : h(x) > Za} (cf., e.g. [11], [14], [26] and [27]). When A = X, we write f O µ instead of f O µ. X
Let f, g : X → Y be measurable functions. We say that functions f|A and g|B are positively dependent with respect to µ and an operator M : µ(F) × µ(F) → µ(F) if for any a, b ∈ Y
µ
f|A > a ∩ g|B > b > µ f| A > a M µ g|B > b ,
(2)
where A, B ∈ F and h|C denotes the restriction of the function h : X → Y to a set C ⊂ X. Obviously, h|C > a = {x ∈ C : h(x) > a} = C ∩ {h > a} . Taking
a M b = min(a, b) and a M b = ab, we recover two important examples of positively dependent functions, namely comonotone functions and independent random variables. A set A ⊂ Y is called monotonically closed if the limit of any nondecreasing sequence of elements from A belongs to A. Among others, the sets Y = [0, 1] and
Y = [0, ∞] posses this property. Throughout the paper we shall assume that φi : Y → Y are arbitrary increasing functions such that φi (Y ) = Y, i = 1, 2, 3, which will imply that φi are continuous. We shall also assume that ? : Y × Y → Y is an arbitrary nondecreasing operator, i.e., a ? c > b ? d whenever a > b and c > d, ♦ : Y × Y → R+ is a nondecreasing function and ◦i : Y × µ(F) → Y, i = 1, 2, 3, are arbitrary functions. Theorem 2.1.
Let Y be monotonically closed. Assume b 7→ a ◦1 b is nondecreasing
for a ∈ Y, and the functions x 7→ x ♦ y and b 7→ a ♦ b are left-continuous for a, y ∈ Y. Assume also that 3
(A1) for any a, b ∈ Y and c, d ∈ µ(F) φ−1 φ1 (a ? b) ◦1 (c M d) > φ−1 φ2 (a) ◦2 c ♦ φ−1 φ3 (b) ◦3 d . 1 2 3
(3)
If f|A and g|B , in which f, g : X → Y and A, B ∈ µ(F), are positively dependent, then φ−1 1
Z
φ1 (f ? g) ◦1 µ
A∩B
>
φ−1 2
Z
φ2 (f ) ◦2 µ
A
♦ φ−1 3
Z
φ3 (g) ◦3 µ .
(4)
B
The positive dependency of f|A and g|B implies that
Proof.
µ A ∩ B ∩ {f > a} ∩ {g > b} > µ A ∩ {f > a} M µ B ∩ {g > b} for a, b ∈ Y. By the monotonicity of ?, f (x)?g(x) > a?b for x ∈ {f > a}∩{g > b} . Hence, by the monotonicity of µ, for a, b ∈ Y
µ A ∩ B ∩ {f ? g > a ? b} > µ A ∩ {f > a} M µ B ∩ {g > b} . As b 7→ a ◦1 b is nondecreasing, φi are increasing and φ1 (Y ) = Y, we have −1 φ1 φ1 (a ? b) ◦1 µ A ∩ B ∩ {φ1 (f ? g) > φ1 (a ? b)} −1 > φ1 φ1 (a ? b) ◦1 µ A ∩ {φ2 (f ) > φ2 (a)} M µ B ∩ {φ3 (g) > φ3 (b)} . From A1 it follows that −1 φ1 φ1 (a ? b) ◦1 µ A ∩ B ∩ {φ1 (f ? g) > φ1 (a ? b)} (5) −1 −1 > φ2 φ2 (a) ◦2 µ A ∩ {φ2 (f ) > φ2 (a)} ♦ φ3 φ3 (b) ◦3 µ B ∩ {φ3 (g) > φ3 (b)} for a, b ∈ Y. Since φ−1 1 is increasing and φ1 (a ? b) ∈ φ1 (Y ) for a, b ∈ Y, Z −1 φ1 (f ? g) ◦1 µ φ1
(6)
A∩B
>
φ−1 2
−1 φ2 (a) ◦2 µ A ∩ {φ2 (f ) > φ2 (a)} ♦ φ3 φ3 (b) ◦3 µ B ∩ {φ3 (g) > φ3 (b)} 4
for a, b ∈ Y, by (5) and the de nition of generalized Sugeno integral. Set H(a) =
a ♦ b and h(a) = φ2 (a) ◦2 µ (A ∩ {φ2 (f ) > φ2 (a)}) . The function Y 3 a 7→ H(φ−1 2 (a)) is nondecreasing and left-continuous as a 7→ a ♦ b and φ−1 2 are nondecreasing and left-continuous. Since Y is monotonically closed and φ2 (Y ) = Y,
−1 sup H φ−1 h(a) 6 H φ sup h(a) , 2 2 a∈Y
(7)
a∈Y
where sup h(a) ∈ Y = φ2 (Y ), by the monotone closedness of Y. Clearly, for any a∈Y
sequence an ∈ Y for which the nondecreasing sequence h(an ) converges to sup h(a), a∈Y
−1 −1 H φ−1 sup h(a) = H φ lim h(a ) = lim H φ h(a ) n n 2 2 2 n→∞ n→∞ a∈Y 6 sup H φ−1 h(a) . 2
(8)
a∈Y
By (7) and (8),
sup H
a∈Y
−1 h(a) = H φ2 sup h(a) .
φ−1 2
(9)
a∈Y
From (6) and (9), for any b ∈ Y Z −1 φ1 φ1 (f ? g) ◦1 µ A∩B
>
φ−1 2
Z
φ2 (f ) ◦2 µ
♦ φ−1 3
φ3 (b) ◦3 µ B ∩ {φ3 (g) > φ3 (b)}
A
as φ2 (Y ) = Y implies that
h i sup h(a) = sup φ2 (a) ◦2 µ A ∩ {φ2 (f ) > φ2 (a)} a∈Y a∈Y Z h i = sup y ◦2 µ A ∩ {φ2 (f ) > y} = φ2 (f ) ◦2 µ. y∈Y A
Proceeding similarly with the supremum in b ∈ Y, we obtain (4).
5
(10)
Remark 1.
(i) The condition A1 is in fact a kind of domination of operators,
which is closely related to the corresponding condition introduced in [16]. (ii) The monotone closedness restriction imposed on Y in the statement of Theorem
2.1 can be replaced by the integrability conditions Z Z φ2 (f ) ◦2 µ ∈ Y and φ1 (f ? g) ◦1 µ ∈ Y,
φ3 (g) ◦3 µ ∈ Y.
(11)
B
A
A∩B
Z
(iii) Theorem 2.1 can be extended to the case of three distinct measures. Replacing the positive dependency condition by µ 1 f |A > a ∩ g |B > b > µ2 f | A > a M µ 3 g |B > b , we can rewrite (4) as Z Z Z −1 −1 −1 φ1 φ1 (f ? g) ◦1 µ1 > φ2 φ2 (f ) ◦2 µ2 ♦ φ3 φ3 (g) ◦3 µ3 . A∩B
A
(12)
B
One can also get an inequality similar to (12) from (4) by measure change. In fact, taking ◦i = ∧, where a ∧ b = min(a, b), we see that for any increasing function φi such that φi (Y ) = Y Z h h i i φi (h) ∧ µ = sup y ∧ µ A ∩ {φi (h) > y} = sup φi (a) ∧ µ A ∩ {h > a} y∈Y
a∈Y
A
Z = φi (S) hdµi , A
where µi (A) = φ−1 µ(A) is a new monotone measure provided φi (0) = 0, and i Z (S) hdµi is the Sugeno integral. Consequently, (4) takes the form A
Z (S)
Z f ? gdµ1 > (S)
A∩B
Z f dµ2 ♦ (S)
A
gdµ3 . B
(iv) We say that fi |A i , i = 1, 2, . . . , n, are positively dependent if for any ai ∈ Y n n n \ o i µ f i |A i > ai > µ {fi > ai } , i=1
i=1
6
where
i
is an n-argument operator on µ(F). Theorem 2.1 can also be easily
extended to the case of n positively dependent functions and ♦ : Y n → R+ . Now we show that, under some additional restrictions, the condition A1 is also necessary to the validity of Chebyshev's inequality (4). Theorem 2.2.
Let Y ⊂ [0, ∞], 0 ∈ Y and 0 M c = c M 0 = 0 for c ∈ µ(F). Let
◦i , i = 1, 2, 3, be such that a ◦i 0 = 0 for a ∈ Y and a 7→ a ◦i b are nondecreasing
for b ∈ µ(F). Assume that (A2) for arbitrary c, d ∈ µ(F) there exist A, B ∈ F such that µ(A) = c, µ(B) = d and µ(A ∩ B) = µ(A) M µ(B). If the Chebyshev inequality (4) is satis ed for all positively dependent functions, then the condition A1 holds true.
Proof.
Fix a, b ∈ Y and c, d ∈ µ(F). De ne f (x) = a and g(x) = b for x ∈
X. Consider A, B ∈ F satisfying A2. Then f|A and g|B are positively dependent because for α > a or β > b we see that
0=µ
f |A > α ∩ g |B > β
>µ
=0 f | A > α M µ g |B > β
as 0 M c = c M 0 = 0, and for α 6 a and β 6 b, we have
µ
f |A > α ∩ g |B > β = µ(A ∩ B) = µ(A) M µ(B) = µ f|A > α M µ g|B > β .
Applying (4) to f|A and g|B yields A1 since Z φ2 (f ) ◦2 µ = max sup y ◦2 µ(A) , sup y ◦2 µ(∅) A
y>φ2 (a)
y6φ2 (a)
= max φ2 (a) ◦2 µ(A), 0 = φ2 (a) ◦2 µ(A), 7
Z
φ3 (g) ◦3 µ = φ3 (b) ◦3 µ(B),
Z
φ1 (f ? g) ◦1 µ = φ1 (a ? b) ◦1 µ(A ∩ B), and
A∩B
A
µ(A ∩ B) = µ(A) M µ(B). The proof is complete.
As a consequence of Theorems 2.1 and 2.2 we have the following result. Theorem 2.3.
Let Y = [0, y¯], where 0 < y¯ 6 ∞, µ(F) = Y, 0 M a = a M 0 = 0
and a ◦i 0 = 0 for a ∈ Y, i = 1, 2, 3. Suppose the functions a 7→ a ◦i b, i = 1, 2, 3, and b 7→ a ◦1 b are nondecreasing while the functions x 7→ x ♦ y and b 7→ a ♦ b are nondecreasing and left-continuous. Suppose also the condition A2 is ful lled. The Chebyshev-type inequality (4) is satis ed for arbitrary positively dependent functions f|A and g|B i the condition A1 holds true. Theorem 2.3 remains valid for Y = [0, y¯) provided that f|A and g|B are integrable in the sense of de nition (11). We now give several examples of pairs of functions which are positively dependent with respect to some operators M . In the sequel, we will assume that
Y = [0, y¯] and M ful lls the assumptions of Theorem 2.3. Example 2.1.
(Comonotone functions) De ne a M b = a ∧ b. The functions f|A
and g|B are positively dependent if for any a, b ∈ Y
µ A ∩ B ∩ {f > a} ∩ {g > b} > µ A ∩ {f > a} ∧ µ B ∩ {g > b} .
(13)
Note that (13) holds if for arbitrary a, b
A ∩ {f > a} ⊂ B ∩ {g > b}
or
B ∩ {g > b} ⊂ A ∩ {f > a} .
(14)
In the case when A = B = X, the functions which satisfy (14) are called comonotone. It is easy to check that functions f, g are comonotone i
f (x) − f (y) g(x) − g(y) > 0 for x, y ∈ Y. 8
(15)
The class of comonotone functions is quite rich. For example, if fi , gi are comonotone for i ∈ I, where I is a nite set, then for any ai , bi > 0 the linear combinations X X ai fi and bi gi are comonotone as well. Moreover, if fi and g are comonotone i∈I
i∈I
for i ∈ I, then sup fi and g are comonotone. The product of nonnegative comonoi∈I
tone functions also posses this property, i.e., if fi and g are comonotone and fi > 0, Y then fi and g are comonotone. The condition A2 is valid for M= ∧ provided i∈I
that for c < d there exist some sets A, B ⊂ Y such that µ(A) = c, µ(B) = d and
A ⊂ B. It is worth pointing out that all the functions f|A and g|B are comonotone if µ is the minitive measure, i.e., µ(A ∩ B) = µ(A) ∧ µ(B) for all A, B (see [26] and
[1]). An example of such a measure is µ(A) = inf m(x), where m is an arbitrary x∈A
nonnegative function on X. Example 2.2.
Set a M b = ab. In the case of probabilistic measures the idea
of positive dependency was introduced by Lehmann [10]. Examples of positively dependent functions include independent random variables. In this case, the condition A2 is satis ed if for arbitrary c, d ∈ Y there exist independent events A, B such that c = µ(A) and d = µ(B). Example 2.3.
Let a M b = (a + b − 1)+ , a, b ∈ Y = [0, 1] be the ukasiewicz
norm. The condition of positive dependency takes the following form µ f|A > a ∩ g|B > b > µ( f|A > a ) + µ( g|B > b ) − 1 , +
and can be rewritten as c c c c 6µ ¯ f |A ≥ a +µ ¯ g |B ≥ b , µ ¯ f|A > a ∪ g|B > b where Ac = [0, 1]\A and µ ¯(A) = 1 − µ(Ac ) is the dual measure provided µ(X) = 1. Observe that if µ ¯ is subadditive, then all functions f|A and g|B are positively dependent with respect to µ. 9
Example 2.4.
Let M be the drastic product on [0, 1], i.e., a M b = a ∧ b when
a = 1 or b = 1, and a M b = 0 when a < 1 and b < 1. If A = X = [0, 1] is the only set for which µ(A) = 1, then arbitrary functions f|A and g|B are positively dependent. Example 2.5.
1/p p p , Consider the Yager product a M b = 1− (1−a) +(1−b) +
p > 0, a, b ∈ [0, 1]. The functions f|A and g|B are positively dependent if for 0 6 a, b 6 1 µp Ac ∪ {f < a} ∪ B c ∪ {g < b} 6 µp Ac ∪ {f < a} + µp B c ∪ {g < b} , where µp (A) = 1 − µ(Ac )
p
is the measure generated by µ. If µp is subadditive,
then all functions are positively dependent. Next we discuss some consequences of Theorem 2.3 concerning the special case
M= ∧. Let Y = [0, y¯], where 0 < y¯ 6 ∞, and let µ(F) = Y. Theorem 2.4.
Assume µ is an arbitrary monotone measure such that µ(X) = Y.
Assume also ◦i , ? and ♦ are nondecreasing operators on Y, a ◦i 0 = 0 = 0 ◦i a for i = 1, 2, 3, and the functions x 7→ x ♦ y and y 7→ x ♦ y are left-continuous. Then,
for arbitrary A, B and positively dependent with respect to the operator ∧ functions f|A : A → Y and g|B : B → Y, the following inequality Z Z Z −1 −1 −1 φ1 φ1 (f ? g) ◦1 µ > φ2 φ2 (f ) ◦2 µ ♦ φ3 φ3 (g) ◦3 µ A∩B
A
(16)
B
holds i for all a, b, c, d ∈ Y φ−1 φ1 (a ? b) ◦1 (c ∧ d) > φ−1 φ2 (a) ◦2 c ♦ φ−1 φ3 (b) ◦3 d . 1 2 3
(17)
Theorem 4.1 of [28] examines the case when ♦ = ?, ◦i = ∧ for i = 1, 2, 3 and
Y = [0, ∞]. If φ1 (x) 6 φi (x) for i = 2, 3 and a ? b 6 a ∧ b for a, b ∈ Y, then (16) holds true. Indeed, from the obvious inequalities
φ−1 φ2 (a) ∧ c ? φ−1 φ3 (b) ∧ d 6 a ? b 2 3 10
(18)
and
−1 −1 −1 φ3 (b) ∧ d 6 φ−1 φ2 (a) ∧ c ? φ−1 φ−1 2 (c) ? φ3 (d) 6 φ1 (c) ∧ φ1 (d), 3 2
(19)
it follows that (17) is true as
φ−1 2
φ2 (a) ∧ c ?
φ−1 3
−1 φ3 (b) ∧ d 6 φ1 φ1 (a ? b) ∧ c ∧ d .
The condition of boundedness from above of the operator ? by the minimum was examined, e.g., in Theorem 3.1 of [1] and Theorem 3.1 of [17]. Ouyang and Mesiar [16] study (16) in the case when ♦ = ?, ◦i = ◦ for all i, Y = [0, 1] and
a ◦ b = T (a, b), where T : [0, 1]2 → [0, 1] is an arbitrary seminorm. They proved that the condition
h i h i (a ? b) ◦ c > (a ◦ c) ? b ∨ a ? (b ◦ c) ,
a, b, c ∈ [0, 1],
(20)
is su cient to the validity of (16) if φi (x) = x for all i. We will show that in this case (20) implies (17). We have
(a ? b) ◦ (c ∧ d) = (a ? b) ◦ c ∧ (a ? b) ◦ d ,
(21)
by the monotonicity of seminorm. From (20) and the property T (x, 1) = x we see that
(a ? b) ◦ c > (a ◦ c) ? b = (a ◦ c) ? (b ◦ 1) > (a ◦ c) ? (b ◦ d)
(22)
(a ? b) ◦ d > a ? (b ◦ d) = (a ◦ 1) ? (b ◦ d) > (a ◦ c) ? (b ◦ d).
(23)
and
Combining (21) with (22) and (23) yields (17), i.e.,
(a ? b) ◦ (c ∧ d) > (a ◦ c) ? (b ◦ d) for all a, b, c, d ∈ [0, 1]. 11
Now we consider the case of M= ∧ and for several choices of operators ?, ♦, ◦i and sets Y provide, by the use of Theorem 2.4, the characterizations of classes of functions φi for which the corresponding inequality (16) is satis ed. 1.
Let Y = [0, y¯], 0 < y¯ 6 ∞, and ♦ = ? = ◦i = ∧ for all i. Then (16) takes the
form
φ−1 1
Z Z Z −1 −1 φ1 (f ∧ g)dµ > φ2 (S) φ2 (f )dµ ∧ φ3 (S) φ3 (g)dµ , (S) B
A
A∩B
(24) where (S)
Z
hdµ = sup y ∧ µ(A ∩ {h > y}) is the classical Sugeno integral. The
A
y∈Y
inequality (24) holds i −1 −1 −1 a ∧ b ∧ φ−1 1 (c) ∧ φ1 (d) > a ∧ φ2 (c) ∧ b ∧ φ3 (d),
a, b, c, d ∈ Y.
(25)
−1 Taking a = b with b = max φ−1 i (c), φi (d) we obtain the following condition i
−1 −1 −1 φ−1 1 (c) ∧ φ1 (d) > φ2 (c) ∧ φ3 (d),
c, d ∈ Y,
which is equivalent to (25). Substituting su ciently large d and c yields φ−1 1 (c) > −1 −1 φ−1 2 (c) and φ1 (d) > φ3 (d). Hence the condition φ1 (c) 6 φi (c), i = 2, 3, is nec-
essary and su cient for the validity of (24). In particular, it holds in the case of
Y = [0, 1], φ1 (x) = x, φ2 (x) = xp and φ3 (x) = xq , where 0 < p, q < 1, corresponding to Hölder's inequality (see [13]). 2.
Let ? = ♦ = · and ◦i = ∧. We will determine all the functions φi for which Z Z Z −1 −1 −1 φ1 (S) φ1 (f g)dµ > φ2 (S) φ2 (f )dµ φ3 (S) φ3 (g)dµ . (26) A∩B
A
B
From Theorem 2.4 it follows that φi have to ful ll the condition
−1 −1 −1 (ab) ∧ φ−1 1 (c) ∧ φ1 (d) > a ∧ φ2 (c) b ∧ φ3 (d) , 12
a, b, c, d ∈ Y.
(27)
We will consider rst the case of Y = [0, 1]. Putting a = b = d = 1 gives φ−1 1 (c) > −1 −1 φ−1 2 (c) for c ∈ Y as φi (Y ) = Y. Taking a = b = c = 1 yields φ1 (d) > φ3 (d) for
d ∈ Y. Hence φ1 (x) 6 φi (x) for i = 2, 3. We claim that the last condition implies (27). To see this note that for any x, y > 0 (28)
1 ∧ (xy) > (1 ∧ x)(1 ∧ y).
Putting x = α/a and y = β/b we obtain that for arbitrary positive a, b and nonnegative α, β (29)
(ab) ∧ (αβ) > (a ∧ α)(b ∧ β). Therefore
−1 −1 −1 (ab) ∧ φ−1 1 (c) ∧ φ1 (d) > (ab) ∧ φ2 (c) ∧ φ3 (d) −1 > (ab) ∧ φ−1 2 (c)φ3 (d) −1 > a ∧ φ−1 (c) b ∧ φ (d) . 2 3 We have thus proved that the condition φ1 (x) 6 φi (x), i = 2, 3, is equivalent to (27). Now we turn to the case of Y = [0, ∞]. Putting a = b = ∞ into (27) we get −1 −1 −1 φ−1 1 (c) ∧ φ1 (d) > φ2 (c)φ3 (d),
c, d ∈ Y.
Since φi (Y ) = Y for i = 1, 2, 3, there are no functions φi satisfying (26) for all comonotone f|A and g|B . 3.
Set Y = [0, 1], ♦ = ? = ◦i = · and φi = φ for all i. We examine for which φ's Z Z Z −1 −1 −1 φ (N ) φ(f g)dµ > φ (N ) φ(f )dµ φ (N ) φ(g)dµ , A∩B
where (N )
Z
A
B
hdµ is the N -integral, called also the Shilkret integral (see [23]). In
A
view of Theorem 2.4 we have to nd all increasing functions φ : Y → Y such that 13
φ(Y ) = Y and φ−1 φ(ab)(c ∧ d) > φ−1 φ(a)c φ−1 φ(b)d ,
a, b, c, d ∈ Y.
(30)
It is easy to check that (30) holds if φ(t) = tp , p > 0. We show that only these functions ful ll (30). Observe that φ(1) = 1. Putting d = 1 into (30) we have
−1 φ(ab)c > φ φ φ(a)c b ,
a, b, c ∈ Y.
Writing x = φ−1 φ(a)c we see that φ(ab) φ(xb) > φ(a) φ(x)
(31)
for a, b, x ∈ Y such that a, x 6= 0. Substituting x = 1 into (31) gives
φ(ab) > φ(a)φ(b),
a, b ∈ [0, 1],
while substituting a = 1 into (31) yields
φ(x)φ(b) > φ(xb),
x, b ∈ [0, 1].
We thus get φ(x)φ(y) = φ(xy), x, y ∈ [0, 1]. The continuity of φ and the condition
φ(1) = 1 implies that φ(x) = xp , p > 0 (see, e.g., [9]). 4.
Assume that Y = [0, 1], ♦ = ? = ◦i = · and φ1 (x) = x. We will determine all the
functions φ2 and φ3 such that the following Hölder type inequality for N -integral Z Z Z −1 −1 f gdµ > φ2 (N ) φ2 (f )dµ φ3 (N ) φ3 (g)dµ , (N ) (32) A∩B
A
B
holds true. The necessary and su cient condition for (32) is of the form
ab(c ∧ d) > φ−1 φ2 (a)c φ−1 φ3 (b)d . 2 3
(33)
Putting b = d = 1 gives
φ2 (ac) > φ2 (a)c, 14
a, c ∈ Y,
(34)
while taking a = c = 1 leads to φ3 (bd) > φ3 (b)d, where b, d ∈ Y. Hence if the functions φi satisfy (33), then
φi (ac) φi (a) > , ac a
0 < a, c 6 1.
(35)
The inequality (35) means that φi (x)/x are nonincreasing on (0, 1]. The conditions (33) and (35) are equivalent as
−1 φ−1 φ (a)c φ φ (b)d 6 (ac)(bd) 6 ab(c ∧ d) 2 3 2 3 for a, b, c, d ∈ Y. We thus have proved that (33) is ful lled for a pair φ2 , φ3 i
φi (x)/x are nonincreasing. Examples of such functions include φi (x) = xpi with 0 < pi < 1 as well as φi (x) = 1.5x1[0,1/2] (x) + 0.5(x + 1)1[1/2,1] (x), in which i = 2, 3. 5.
The reverse Minkowski inequality Z Z Z −1 −1 −1 φ φ(f + g) ◦ µ > φ φ(f ) ◦ µ + φ φ(g) ◦ µ
is not valid for the Sugeno integral (◦ = ∧), the N -integral (◦ = ·) and for any other integrals with respect to the operators ◦ such that a ◦ 0 = 0 and a ◦ 1 > 0 for a > 0. To see this observe that putting c = 0 into the inequality −1
φ
−1 −1 φ(a + b) ◦ (c ∧ d) > φ φ(a) ◦ c + φ φ(b) ◦ d
we get the condition 0 > φ(b) ◦ d for b, d ∈ Y, which is not satis ed for any increasing function provided 1 ∈ Y. The above examples concern the case of comonotone functions f and g corresponding to M= ∧, which is the only one which has been studied previously. Now 15
we will consider the case when functions f and g are positively dependent with respect to the product operator. 6.
Let M= ·, ? = ♦ = · and ◦i = ∧. Our goal is to determine functions φi for
which
φ−1 1
Z Z Z −1 −1 φ1 (f g)dµ > φ2 (S) φ2 (f )dµ φ3 (S) φ3 (g)dµ , (S) B
A
A∩B
(36)
where f|A and g|B are positively dependent with respect to ·. To do that we only need to nd the functions φi ful lling the inequality
−1 −1 (ab) ∧ φ−1 (cd) > a ∧ φ (c) b ∧ φ (d) , 1 2 3
(37)
where a, b, c, d ∈ Y = [0, 1]. Writing a = b = 1 we have −1 −1 φ−1 1 (cd) > φ2 (c)φ3 (d),
c, d ∈ [0, 1].
(38)
We will show the equivalency of (37) and (38). If (38) holds, then
−1 −1 −1 −1 (ab) ∧ φ−1 1 (cd) > (ab) ∧ φ2 (c)φ3 (d) > a ∧ φ2 (c) b ∧ φ3 (d) , by (29), so (37) is satis ed. Substituting c = φ2 (x) and d = φ3 (y) into (38) leads to the equivalent condition:
φ2 (x)φ3 (y) > φ1 (xy),
x, y ∈ [0, 1].
(39)
Thus we have showed that (36) is valid for all positively dependent functions f|A and g|B i the functions φi ful ll the condition (39). For example, φi (x) = xp , where
p > 0, i = 1, 2, 3, as well as φ1 (x) = x and arbitrary increasing functions φ2 and φ3 such that φi (x)/x is nonincreasing and φi [0, 1] = [0, 1], meet the condition (39). The latter statement is a consequence of the fact that φ2 (x)/x φ3 (y)/y > φ2 (1)φ3 (1) = 1. 16
7.
Let M be the ukasiewicz norm on [0, 1] (see Example 2.3), ◦i = · for all i and
? = ♦. In the case of φi (x) = x, i = 1, 2, 3, the necessary and su cient condition to the validity of (4) for positively dependent with respect to M functions f and g is
(a ? b)(c + d − 1)+ > (ac) ? (bd),
a, b, c, d ∈ [0, 1].
(40)
We show that the continuous operator a ? b = (a + b − 1)+ ful lls (40). Obviously, (40) holds true if ac + bd < 1. Suppose ac + bd ≥ 1. Since a + b > ac + bd > 1 and c + d > ac + bd > 1, we see that for ac + bd > 1, (40) takes the form
(a + b − 1)(c + d − 1) > ac + bd − 1. The last inequality is true because it is equivalent to the following one: (a − 1)(d − 1) + (b − 1)(c − 1) > 0. An interesting open problem is to characterize the class of left-continuous operators ? satisfying the condition (40).
3
Related results
Throughout this section we shall use the notation and work under the assumptions introduced before the statement of Theorem 2.1. The following theorem gives the su cient conditions to the validity of the reverse Chebyshev-type inequality, i.e., the inequality opposite to (4). Theorem 3.1.
Assume Y is monotonically closed, b 7→ a ◦1 b is nondecreasing for
a ∈ Y, ? is an increasing operator and Y ? Y = Y, where Y ? Y = {a ? b : a, b ∈ Y } .
Assume also that (B1) for a xed A ∈ F and for any a, b ∈ Y and M : µ(F) × µ(F) → µ(F) µ
f|A > a ∪ g|A > b 6 µ f|A > a M µ g|A > b ,
17
(41)
(B2) for any a, b ∈ Y and c, d ∈ µ(F) φ−1 φ1 (a ? b) ◦1 (c M d) 6 φ−1 φ2 (a) ◦2 c ♦ φ−1 φ3 (b) ◦3 d . 1 2 3
Then φ−1 1
Z
φ1 (f ? g) ◦1 µ
A
Proof.
6
φ−1 2
Z
φ2 (f ) ◦2 µ
A
♦ φ−1 3
Z
φ3 (g) ◦3 µ .
(42)
A
Since ? is increasing, we have that f (x) ? g(x) < a ? b if f (x) < a and
g(x) < b. Hence for any A A ∩ {f ? g > a ? b} ⊂ A ∩ {f > a} ∪ {g > b} = A ∩ {f > a} ∪ A ∩ {g > b} . (43) From B1 we conclude that for a, b ∈ Y φ−1 φ (a ? b) ◦ µ A ∩ {φ (f ? g) > φ (a ? b)} 1 1 1 1 1 6 φ−1 φ (a ? b) ◦ µ(A ∩ {φ (f ) > φ (a)}) M µ(A ∩ {φ (g) > φ (b)}) . 1 1 2 2 3 3 1 From B2 we obtain −1 φ1 φ1 (a ? b) ◦1 µ A ∩ {φ1 (f ? g) > φ1 (a ? b)} −1 −1 6 φ2 φ2 (a) ◦2 µ {A ∩ φ2 (f ) > φ2 (a)} ♦ φ3 φ3 (b) ◦3 µ {A ∩ φ3 (g) > φ3 (b)} . (44) Taking the supremums with respect to a ∈ Y and b ∈ Y of the right-hand side of (44) rst and of the left-hand side of (44) afterwards completes the proof.
Example 3.1.
Let M= ∨, where a ∨ b = max(a, b), and let Y = [0, 1]. If f and g
are comonotone, then B1 is satis ed for any A ∈ F, and the inequality Z Z Z −1 −1 −1 φ1 (N ) φ1 (f g)dµ 6 φ2 (N ) φ2 (f )dµ ∨ φ3 (N ) φ3 (g)dµ , (45) A
A
A
18
holds true provided that
φ−1 φ1 (ab)(c ∨ d) 6 φ−1 φ2 (a)c ∨ φ−1 φ3 (b)d 1 2 3
(46)
φ3 (b)c > φ2 (ab)c and φ−1 φ2 (a)c > φ−1 for all a, b, c, d ∈ [0, 1]. Since φ−1 3 2 2 φ3 (ab)d , the condition (46) is ful lled if φ−1 3 φi (x)y , φ−1 φ1 (x)y 6 φ−1 1 i
x, y ∈ [0, 1], i = 2, 3.
For example, (45) is valid for φi (x) = φ(x), i = 1, 2, 3, where φ is an increasing function, and for φi (x) = xpi , where 0 < p1 6 p2 6 p3 . It is worth to note that B1 holds for all functions f, g if M= ∨ and µ is the maxitive measure, i.e., µ(A ∪ B) = max µ(A), µ(B) for all A, B. An example of such a measure is the possibility measure µ(A) = sup m(x), where m : X → [0, ∞] is an arbitrary function such that µ(X) > 0.
x∈A
Wu et al. [28] proved that if a ? b > a ∨ b for a, b ∈ [0, ∞] and φ1 (x) > φi (x),
i = 2, 3, then for any comonotone functions f, g Z Z Z −1 −1 −1 φ1 (S) φ1 (f ? g)dµ 6 φ2 (S) φ2 (f )dµ ? φ3 (S) φ3 (g)dµ . (47) A
A
A
Our su cient condition for (47) is of the form
φ−1 φ1 (a ? b) ∧ (c ∨ d) 6 φ−1 φ2 (a) ∧ c ? φ−1 φ3 (b) ∧ d . 1 2 3
(48)
Note that none of the conditions a ? b > a ∨ b and (48) does not imply the other. By the monotonicity of the Sugeno integral, we have Z Z Z (S) f ∧ gdµ 6 (S) f dµ ∧ (S) gdµ A
A
(49)
A
for any f, g and A ∈ F (see [26]). Unfortunately, (49) does not follow neither from Theorem 3.1 and nor from [28] since the operator ? = ∧ is not bounded from below 19
by the maximum. Moreover, the inequality (49) does not follow from Theorem 3.1, even if f and g are comonotone because the condition B2 of the form
(a ∧ b) ∧ (c ∨ d) 6 (a ∧ c) ∧ (b ∧ d),
a, b, c, d > 0
is not ful lled for all a > b > c > d. This means that even in the case of the classical Sugeno integral, the condition B2 is not necessary for (42). Next theorem provides, under some additional restrictions on the operators M and ◦i , the necessary conditions to the validity of (42). Theorem 3.2.
Let Y = [0, y¯] for 0 < y¯ 6 ∞. Suppose that a M 0 > a, 0 M
a > a and a M a > a for a ∈ Y. Suppose also that the functions a 7→ a ◦i b are
nondecreasing and a ◦i 0 = 0 for a ∈ Y, i = 1, 2, 3. If (42) is valid for f, g satisfying (41),
then
(B3) for a, b ∈ Y and c ∈ µ(F) φ−1 φ1 (a ? b) ◦1 c 6 φ−1 φ2 (a) ◦2 c ♦ φ−1 φ3 (b) ◦3 c . 1 2 3
Proof.
Let α, β ∈ Y and A be an arbitrary measurable set. Set f (x) = α and
g(x) = β for all x ∈ X. It is easy to verify that f|A and g|A ful ll the condition B1, Z Z Z φ1 (f ? g) ◦1 µ = φ1 (α ? β) ◦1 µ(A), φ2 (f ) ◦2 µ = φ2 (α) ◦2 µ(A), φ3 (g) ◦3 µ = φ3 (β) ◦3 µ(A). A
A
A
Hence B3 follows from (42).
Example 3.2.
Let M= ∨. It is not true that Z 1 Z 1 Z (S) f gdµ 6 (S) f dµ (S) 0
0
1
gdµ
0
as the inequality B3 of the form (ab)∧c 6 (a∧c)(b∧c) does not hold for a = b = 1. 20
The operators M= ∨ and M= + meet the assumptions of Theorem 3.2 for any
Y, the operator a M b = 1 − (1 − a)(1 − b) meets these assumptions for Y = [0, 1], while the operator M= · does not meet them. We do not know whether there are operators M, ?, ♦, ◦i for which the condition B3 is necessary and su cient. As an open problem we propose to establish the necessary and su cient conditions under which the inequality (42) holds for all functions satisfying (41).
4
Conclusions
We have presented the necessary and su cient conditions to the validity of Chebyshev type and reverse Chebyshev type inequalities for generalized Sugeno integrals in the case of functions belonging to an essentially wider class than the comonotone functions. For speci c choices of operators, we have characterized classes of functions for which the Chebyshev type inequality for the classical Sugeno integral is satis ed. The proofs consist in solving some functional inequalities.
5
Acknowledgments
The authors would like to thank the referees for their valuable comments which led to improvements in the paper. The work on this paper was partially supported by the grant for young researchers from Lodz University of Technology I-2 501 plan 2 number 12/2013.
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H. Agahi, R. Mesiar, Y. Ouyang, On some advanced type inequalities for Sugeno integral and T-(S-)evaluators, Information Sciences 190 (2012) 64-75.
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J. Caballero, K. Sadarangani, Chebyshev inequality for Sugeno integrals, Fuzzy Sets and Systems 161 (2010) 1480-1487.
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T. Chen, H. Chang, G. Tzeng, Using fuzzy measures and habitual domains to analyze the public attitude and apply to the gas taxi policy, European Journal of Operational Research 137 (2002) 145-161.
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A. Flores-Franuli£, H. Román-Flores, A Chebyshev type inequality for fuzzy integrals, Applied Mathematics and Computation 190 (2007) 1178-1184.
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Y. Hu, Fuzzy integral-based perceptron for two-class pattern classi cation problems, Information Sciences 177 (2007) 1673-1686.
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M. Kuczma, An Introduction to the Theory of Functional Equations and Inequalities. Cauchy's Equation and Jensen's Inequality, Birkhäuser, Basel, 2009.
[10] E.L. Lehmann, Some concepts of dependence, Annals Mathematics Statistics Volume 37, Number 5 (1966) 1137-1153. [11] R. Mesiar, Fuzzy measures and integrals, Fuzzy Sets and Systems 156 (2005) 365-370.
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[12] R. Mesiar, Y. Ouyang, General Chebyshev type inequalities for Sugeno integrals, Fuzzy Sets and Systems 160 (2009) 58-64. [13] D.S. Mitrinovi¢, J.E. Pe£ari¢, A.M. Fink, Classical and New Inequalities in Analysis, Kluwer Academic Publisher, 1993. [14] T. Murofushi, M. Sugeno, Fuzzy t-conorm integral with respect to fuzzy measures: Generalization of Sugeno integral and Choquet integral, Fuzzy Sets and Systems 42 (1991) 57-71. [15] Y. Narukawa, V. Torra, Fuzzy measures and integrals in evaluation of strategies, Information Sciences 177 (2007) 4686-4695. [16] Y. Ouyang, R. Mesiar, On the Chebyshev type inequality for seminormed fuzzy integral, Applied Mathematics Letters 22 (2009) 1810-1815. [17] Y. Ouyang, R. Mesiar, H. Agahi, An inequality related to Minkowski type for Sugeno integrals, Information Sciences 180 (2010) 27932801. [18] D. Ralescu, G. Adams, The fuzzy integral, Journal of Mathematical Analysis and Applications 75 (2) 562-570. [19] H. Román-Flores, A. Flores-Franuli£, R. Bassanezi, M. Rojas-Medar, On the level-continuity of fuzzy integrals, Fuzzy Sets and Systems 80 (1996) 339-344. [20] H. Román-Flores, Y. Chalco-Cano, H-Continuity of fuzzy measures and set defuzzi cation, Fuzzy Sets and Systems 157 (2006) 230-242. [21] H. Román-Flores, A. Flores-Franuli£, Y. Chalco-Cano, The fuzzy integral for monotone functions, Applied Mathematics and Computation 185 (2007) 492498.
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[22] H. Román-Flores, A. Flores-Franuli£, Y. Chalco-Cano, A Jensen type inequality for fuzzy integrals, Information Sciences 177 (2007) 3192-3201. [23] N. Shilkret, Maxitive measure and integration, Indagationes Mathematicae 33 (1971) 109-116. [24] M. Sugeno, Theory of Fuzzy Integrals and its Applications, Ph.D. Dissertation, Tokyo Institute of Technology, 1974. [25] V. Torra, Y. Narukawa, The h-index and the number of citations: Two fuzzy integrals, IEEE Transactions on Fuzzy Systems 16 (2008) 795-797. [26] Z. Wang, G. Klir, Fuzzy Measure Theory, Plenum, New York, 1992. [27] C. Wu, S. Wang, M. Ma, Generalized fuzzy integrals: Part I. Fundamental concepts, Fuzzy Sets and Systems 57 (1993) 219-226. [28] L. Wu, J. Sun, X. Ye, L. Zhu, Hölder type inequality for Sugeno integral, Fuzzy Sets and Systems 161 (2010) 2337-2347.
MAREK KALUSZKA INSTITUTE OF MATHEMATICS LODZ UNIVERSITY OF TECHNOLOGY UL. WÓLCZASKA 215 90-924 LODZ POLAND E-MAIL: [email protected] ANDRZEJ OKOLEWSKI INSTITUTE OF MATHEMATICS LODZ UNIVERSITY OF TECHNOLOGY UL. WÓLCZASKA 215 24
90-924 LODZ POLAND E-MAIL: [email protected] MICHAL BOCZEK INSTITUTE OF MATHEMATICS LODZ UNIVERSITY OF TECHNOLOGY UL. WÓLCZASKA 215 90-924 LODZ POLAND E-MAIL: [email protected]
25
Abstract We give the necessary and su cient conditions guaranteeing the validity of Chebyshev type inequalities for generalized Sugeno integrals in the case of functions belonging to a much wider class than the comonotone functions. For several choices of operators, we characterize the classes of functions for which the Chebyshev type inequality for the classical Sugeno integral is satis ed.
1
Introduction
The pioneering concept of the fuzzy integral was introduced by Sugeno [24] as a tool for modeling non-deterministic problems. Theoretical investigations of the integral and its generalizations have been pursued by many researchers. Among others, Ralescu and Adams [18] provided several characterizations of the Sugeno integral and proved a monotone convergence theorem for this integral, RománFlores et al. [19, 20] discussed level-continuity of fuzzy integrals and H-continuity of fuzzy measures, while Wang and Klir [26] presented an excellent general overview on fuzzy measurement and fuzzy integration theory. On the other hand, fuzzy integrals have also been successfully applied to various elds (see, e.g., [6], [8],
[15], [25]).
Corresponding author. E-mail adress: [email protected]; tel.: +48 42 6313859; fax.: +48 42 6363114. 1
1
The study of inequalities for Sugeno integral was initiated by Román-Flores et al. [21]. Since then, the fuzzy integral counterparts of several classical inequalities, including Chebyshev's, Jensen's, Minkowski's and Hölder's inequalities, are given by Flores-Franuli¢ and Román-Flores [7], Agahi et al. [2, 3, 4], Mesiar and Ouyang
[12], Román-Flores et al. [22], and other researchers. The purpose of this paper is to study not only the su cient conditions for the validity of Chebyshev type and reverse Chebyshev type inequalities for the generalized Sugeno integral (cf. [1], [3], [5],[7], [12] and [16]), but also the necessary ones. The results are obtained for quite universal integrals and a rich class of functions, including the comonotone functions as a special case. For some speci c choices of operators, the corresponding characterizations of classes of functions for which Chebyshev's inequality for the classical Sugeno integral is satis ed are also presented. The paper is organized as follows. In Sections 2 and 3 we set up notation and terminology and present our main results, some related results as well as several illustrative examples. Concluding remarks are given in Section 4.
2
Main results
Let (X, F) be a measurable space and µ : F → [0, ∞] be a monotone measure, i.e., µ(∅) = 0, µ(X) > 0 and µ(A) 6 µ(B) whenever A ⊂ B. In the literature monotone measures are also called fuzzy measures (see, e.g., [26]). Let Y ⊂ R be an arbitrary nonempty interval; usually Y = [0, 1], Y = [0, ∞], Y = [0, ∞) or
Y = R. We denote the range of µ by µ(F). For a measurable function h : X → Y, we will de ne the Sugeno integral of h on a set A ∈ F with respect to µ and an
2
operator O : Y × µ(F) → Y as Z h O µ = sup a O µ(A ∩ {h > a}) ,
(1)
a∈Y A
where {h > a} stands for Z {x ∈ X : h(x) > Za} (cf., e.g. [11], [14], [26] and [27]). When A = X, we write f O µ instead of f O µ. X
Let f, g : X → Y be measurable functions. We say that functions f|A and g|B are positively dependent with respect to µ and an operator M : µ(F) × µ(F) → µ(F) if for any a, b ∈ Y
µ
f|A > a ∩ g|B > b > µ f| A > a M µ g|B > b ,
(2)
where A, B ∈ F and h|C denotes the restriction of the function h : X → Y to a set C ⊂ X. Obviously, h|C > a = {x ∈ C : h(x) > a} = C ∩ {h > a} . Taking
a M b = min(a, b) and a M b = ab, we recover two important examples of positively dependent functions, namely comonotone functions and independent random variables. A set A ⊂ Y is called monotonically closed if the limit of any nondecreasing sequence of elements from A belongs to A. Among others, the sets Y = [0, 1] and
Y = [0, ∞] posses this property. Throughout the paper we shall assume that φi : Y → Y are arbitrary increasing functions such that φi (Y ) = Y, i = 1, 2, 3, which will imply that φi are continuous. We shall also assume that ? : Y × Y → Y is an arbitrary nondecreasing operator, i.e., a ? c > b ? d whenever a > b and c > d, ♦ : Y × Y → R+ is a nondecreasing function and ◦i : Y × µ(F) → Y, i = 1, 2, 3, are arbitrary functions. Theorem 2.1.
Let Y be monotonically closed. Assume b 7→ a ◦1 b is nondecreasing
for a ∈ Y, and the functions x 7→ x ♦ y and b 7→ a ♦ b are left-continuous for a, y ∈ Y. Assume also that 3
(A1) for any a, b ∈ Y and c, d ∈ µ(F) φ−1 φ1 (a ? b) ◦1 (c M d) > φ−1 φ2 (a) ◦2 c ♦ φ−1 φ3 (b) ◦3 d . 1 2 3
(3)
If f|A and g|B , in which f, g : X → Y and A, B ∈ µ(F), are positively dependent, then φ−1 1
Z
φ1 (f ? g) ◦1 µ
A∩B
>
φ−1 2
Z
φ2 (f ) ◦2 µ
A
♦ φ−1 3
Z
φ3 (g) ◦3 µ .
(4)
B
The positive dependency of f|A and g|B implies that
Proof.
µ A ∩ B ∩ {f > a} ∩ {g > b} > µ A ∩ {f > a} M µ B ∩ {g > b} for a, b ∈ Y. By the monotonicity of ?, f (x)?g(x) > a?b for x ∈ {f > a}∩{g > b} . Hence, by the monotonicity of µ, for a, b ∈ Y
µ A ∩ B ∩ {f ? g > a ? b} > µ A ∩ {f > a} M µ B ∩ {g > b} . As b 7→ a ◦1 b is nondecreasing, φi are increasing and φ1 (Y ) = Y, we have −1 φ1 φ1 (a ? b) ◦1 µ A ∩ B ∩ {φ1 (f ? g) > φ1 (a ? b)} −1 > φ1 φ1 (a ? b) ◦1 µ A ∩ {φ2 (f ) > φ2 (a)} M µ B ∩ {φ3 (g) > φ3 (b)} . From A1 it follows that −1 φ1 φ1 (a ? b) ◦1 µ A ∩ B ∩ {φ1 (f ? g) > φ1 (a ? b)} (5) −1 −1 > φ2 φ2 (a) ◦2 µ A ∩ {φ2 (f ) > φ2 (a)} ♦ φ3 φ3 (b) ◦3 µ B ∩ {φ3 (g) > φ3 (b)} for a, b ∈ Y. Since φ−1 1 is increasing and φ1 (a ? b) ∈ φ1 (Y ) for a, b ∈ Y, Z −1 φ1 (f ? g) ◦1 µ φ1
(6)
A∩B
>
φ−1 2
−1 φ2 (a) ◦2 µ A ∩ {φ2 (f ) > φ2 (a)} ♦ φ3 φ3 (b) ◦3 µ B ∩ {φ3 (g) > φ3 (b)} 4
for a, b ∈ Y, by (5) and the de nition of generalized Sugeno integral. Set H(a) =
a ♦ b and h(a) = φ2 (a) ◦2 µ (A ∩ {φ2 (f ) > φ2 (a)}) . The function Y 3 a 7→ H(φ−1 2 (a)) is nondecreasing and left-continuous as a 7→ a ♦ b and φ−1 2 are nondecreasing and left-continuous. Since Y is monotonically closed and φ2 (Y ) = Y,
−1 sup H φ−1 h(a) 6 H φ sup h(a) , 2 2 a∈Y
(7)
a∈Y
where sup h(a) ∈ Y = φ2 (Y ), by the monotone closedness of Y. Clearly, for any a∈Y
sequence an ∈ Y for which the nondecreasing sequence h(an ) converges to sup h(a), a∈Y
−1 −1 H φ−1 sup h(a) = H φ lim h(a ) = lim H φ h(a ) n n 2 2 2 n→∞ n→∞ a∈Y 6 sup H φ−1 h(a) . 2
(8)
a∈Y
By (7) and (8),
sup H
a∈Y
−1 h(a) = H φ2 sup h(a) .
φ−1 2
(9)
a∈Y
From (6) and (9), for any b ∈ Y Z −1 φ1 φ1 (f ? g) ◦1 µ A∩B
>
φ−1 2
Z
φ2 (f ) ◦2 µ
♦ φ−1 3
φ3 (b) ◦3 µ B ∩ {φ3 (g) > φ3 (b)}
A
as φ2 (Y ) = Y implies that
h i sup h(a) = sup φ2 (a) ◦2 µ A ∩ {φ2 (f ) > φ2 (a)} a∈Y a∈Y Z h i = sup y ◦2 µ A ∩ {φ2 (f ) > y} = φ2 (f ) ◦2 µ. y∈Y A
Proceeding similarly with the supremum in b ∈ Y, we obtain (4).
5
(10)
Remark 1.
(i) The condition A1 is in fact a kind of domination of operators,
which is closely related to the corresponding condition introduced in [16]. (ii) The monotone closedness restriction imposed on Y in the statement of Theorem
2.1 can be replaced by the integrability conditions Z Z φ2 (f ) ◦2 µ ∈ Y and φ1 (f ? g) ◦1 µ ∈ Y,
φ3 (g) ◦3 µ ∈ Y.
(11)
B
A
A∩B
Z
(iii) Theorem 2.1 can be extended to the case of three distinct measures. Replacing the positive dependency condition by µ 1 f |A > a ∩ g |B > b > µ2 f | A > a M µ 3 g |B > b , we can rewrite (4) as Z Z Z −1 −1 −1 φ1 φ1 (f ? g) ◦1 µ1 > φ2 φ2 (f ) ◦2 µ2 ♦ φ3 φ3 (g) ◦3 µ3 . A∩B
A
(12)
B
One can also get an inequality similar to (12) from (4) by measure change. In fact, taking ◦i = ∧, where a ∧ b = min(a, b), we see that for any increasing function φi such that φi (Y ) = Y Z h h i i φi (h) ∧ µ = sup y ∧ µ A ∩ {φi (h) > y} = sup φi (a) ∧ µ A ∩ {h > a} y∈Y
a∈Y
A
Z = φi (S) hdµi , A
where µi (A) = φ−1 µ(A) is a new monotone measure provided φi (0) = 0, and i Z (S) hdµi is the Sugeno integral. Consequently, (4) takes the form A
Z (S)
Z f ? gdµ1 > (S)
A∩B
Z f dµ2 ♦ (S)
A
gdµ3 . B
(iv) We say that fi |A i , i = 1, 2, . . . , n, are positively dependent if for any ai ∈ Y n n n \ o i µ f i |A i > ai > µ {fi > ai } , i=1
i=1
6
where
i
is an n-argument operator on µ(F). Theorem 2.1 can also be easily
extended to the case of n positively dependent functions and ♦ : Y n → R+ . Now we show that, under some additional restrictions, the condition A1 is also necessary to the validity of Chebyshev's inequality (4). Theorem 2.2.
Let Y ⊂ [0, ∞], 0 ∈ Y and 0 M c = c M 0 = 0 for c ∈ µ(F). Let
◦i , i = 1, 2, 3, be such that a ◦i 0 = 0 for a ∈ Y and a 7→ a ◦i b are nondecreasing
for b ∈ µ(F). Assume that (A2) for arbitrary c, d ∈ µ(F) there exist A, B ∈ F such that µ(A) = c, µ(B) = d and µ(A ∩ B) = µ(A) M µ(B). If the Chebyshev inequality (4) is satis ed for all positively dependent functions, then the condition A1 holds true.
Proof.
Fix a, b ∈ Y and c, d ∈ µ(F). De ne f (x) = a and g(x) = b for x ∈
X. Consider A, B ∈ F satisfying A2. Then f|A and g|B are positively dependent because for α > a or β > b we see that
0=µ
f |A > α ∩ g |B > β
>µ
=0 f | A > α M µ g |B > β
as 0 M c = c M 0 = 0, and for α 6 a and β 6 b, we have
µ
f |A > α ∩ g |B > β = µ(A ∩ B) = µ(A) M µ(B) = µ f|A > α M µ g|B > β .
Applying (4) to f|A and g|B yields A1 since Z φ2 (f ) ◦2 µ = max sup y ◦2 µ(A) , sup y ◦2 µ(∅) A
y>φ2 (a)
y6φ2 (a)
= max φ2 (a) ◦2 µ(A), 0 = φ2 (a) ◦2 µ(A), 7
Z
φ3 (g) ◦3 µ = φ3 (b) ◦3 µ(B),
Z
φ1 (f ? g) ◦1 µ = φ1 (a ? b) ◦1 µ(A ∩ B), and
A∩B
A
µ(A ∩ B) = µ(A) M µ(B). The proof is complete.
As a consequence of Theorems 2.1 and 2.2 we have the following result. Theorem 2.3.
Let Y = [0, y¯], where 0 < y¯ 6 ∞, µ(F) = Y, 0 M a = a M 0 = 0
and a ◦i 0 = 0 for a ∈ Y, i = 1, 2, 3. Suppose the functions a 7→ a ◦i b, i = 1, 2, 3, and b 7→ a ◦1 b are nondecreasing while the functions x 7→ x ♦ y and b 7→ a ♦ b are nondecreasing and left-continuous. Suppose also the condition A2 is ful lled. The Chebyshev-type inequality (4) is satis ed for arbitrary positively dependent functions f|A and g|B i the condition A1 holds true. Theorem 2.3 remains valid for Y = [0, y¯) provided that f|A and g|B are integrable in the sense of de nition (11). We now give several examples of pairs of functions which are positively dependent with respect to some operators M . In the sequel, we will assume that
Y = [0, y¯] and M ful lls the assumptions of Theorem 2.3. Example 2.1.
(Comonotone functions) De ne a M b = a ∧ b. The functions f|A
and g|B are positively dependent if for any a, b ∈ Y
µ A ∩ B ∩ {f > a} ∩ {g > b} > µ A ∩ {f > a} ∧ µ B ∩ {g > b} .
(13)
Note that (13) holds if for arbitrary a, b
A ∩ {f > a} ⊂ B ∩ {g > b}
or
B ∩ {g > b} ⊂ A ∩ {f > a} .
(14)
In the case when A = B = X, the functions which satisfy (14) are called comonotone. It is easy to check that functions f, g are comonotone i
f (x) − f (y) g(x) − g(y) > 0 for x, y ∈ Y. 8
(15)
The class of comonotone functions is quite rich. For example, if fi , gi are comonotone for i ∈ I, where I is a nite set, then for any ai , bi > 0 the linear combinations X X ai fi and bi gi are comonotone as well. Moreover, if fi and g are comonotone i∈I
i∈I
for i ∈ I, then sup fi and g are comonotone. The product of nonnegative comonoi∈I
tone functions also posses this property, i.e., if fi and g are comonotone and fi > 0, Y then fi and g are comonotone. The condition A2 is valid for M= ∧ provided i∈I
that for c < d there exist some sets A, B ⊂ Y such that µ(A) = c, µ(B) = d and
A ⊂ B. It is worth pointing out that all the functions f|A and g|B are comonotone if µ is the minitive measure, i.e., µ(A ∩ B) = µ(A) ∧ µ(B) for all A, B (see [26] and
[1]). An example of such a measure is µ(A) = inf m(x), where m is an arbitrary x∈A
nonnegative function on X. Example 2.2.
Set a M b = ab. In the case of probabilistic measures the idea
of positive dependency was introduced by Lehmann [10]. Examples of positively dependent functions include independent random variables. In this case, the condition A2 is satis ed if for arbitrary c, d ∈ Y there exist independent events A, B such that c = µ(A) and d = µ(B). Example 2.3.
Let a M b = (a + b − 1)+ , a, b ∈ Y = [0, 1] be the ukasiewicz
norm. The condition of positive dependency takes the following form µ f|A > a ∩ g|B > b > µ( f|A > a ) + µ( g|B > b ) − 1 , +
and can be rewritten as c c c c 6µ ¯ f |A ≥ a +µ ¯ g |B ≥ b , µ ¯ f|A > a ∪ g|B > b where Ac = [0, 1]\A and µ ¯(A) = 1 − µ(Ac ) is the dual measure provided µ(X) = 1. Observe that if µ ¯ is subadditive, then all functions f|A and g|B are positively dependent with respect to µ. 9
Example 2.4.
Let M be the drastic product on [0, 1], i.e., a M b = a ∧ b when
a = 1 or b = 1, and a M b = 0 when a < 1 and b < 1. If A = X = [0, 1] is the only set for which µ(A) = 1, then arbitrary functions f|A and g|B are positively dependent. Example 2.5.
1/p p p , Consider the Yager product a M b = 1− (1−a) +(1−b) +
p > 0, a, b ∈ [0, 1]. The functions f|A and g|B are positively dependent if for 0 6 a, b 6 1 µp Ac ∪ {f < a} ∪ B c ∪ {g < b} 6 µp Ac ∪ {f < a} + µp B c ∪ {g < b} , where µp (A) = 1 − µ(Ac )
p
is the measure generated by µ. If µp is subadditive,
then all functions are positively dependent. Next we discuss some consequences of Theorem 2.3 concerning the special case
M= ∧. Let Y = [0, y¯], where 0 < y¯ 6 ∞, and let µ(F) = Y. Theorem 2.4.
Assume µ is an arbitrary monotone measure such that µ(X) = Y.
Assume also ◦i , ? and ♦ are nondecreasing operators on Y, a ◦i 0 = 0 = 0 ◦i a for i = 1, 2, 3, and the functions x 7→ x ♦ y and y 7→ x ♦ y are left-continuous. Then,
for arbitrary A, B and positively dependent with respect to the operator ∧ functions f|A : A → Y and g|B : B → Y, the following inequality Z Z Z −1 −1 −1 φ1 φ1 (f ? g) ◦1 µ > φ2 φ2 (f ) ◦2 µ ♦ φ3 φ3 (g) ◦3 µ A∩B
A
(16)
B
holds i for all a, b, c, d ∈ Y φ−1 φ1 (a ? b) ◦1 (c ∧ d) > φ−1 φ2 (a) ◦2 c ♦ φ−1 φ3 (b) ◦3 d . 1 2 3
(17)
Theorem 4.1 of [28] examines the case when ♦ = ?, ◦i = ∧ for i = 1, 2, 3 and
Y = [0, ∞]. If φ1 (x) 6 φi (x) for i = 2, 3 and a ? b 6 a ∧ b for a, b ∈ Y, then (16) holds true. Indeed, from the obvious inequalities
φ−1 φ2 (a) ∧ c ? φ−1 φ3 (b) ∧ d 6 a ? b 2 3 10
(18)
and
−1 −1 −1 φ3 (b) ∧ d 6 φ−1 φ2 (a) ∧ c ? φ−1 φ−1 2 (c) ? φ3 (d) 6 φ1 (c) ∧ φ1 (d), 3 2
(19)
it follows that (17) is true as
φ−1 2
φ2 (a) ∧ c ?
φ−1 3
−1 φ3 (b) ∧ d 6 φ1 φ1 (a ? b) ∧ c ∧ d .
The condition of boundedness from above of the operator ? by the minimum was examined, e.g., in Theorem 3.1 of [1] and Theorem 3.1 of [17]. Ouyang and Mesiar [16] study (16) in the case when ♦ = ?, ◦i = ◦ for all i, Y = [0, 1] and
a ◦ b = T (a, b), where T : [0, 1]2 → [0, 1] is an arbitrary seminorm. They proved that the condition
h i h i (a ? b) ◦ c > (a ◦ c) ? b ∨ a ? (b ◦ c) ,
a, b, c ∈ [0, 1],
(20)
is su cient to the validity of (16) if φi (x) = x for all i. We will show that in this case (20) implies (17). We have
(a ? b) ◦ (c ∧ d) = (a ? b) ◦ c ∧ (a ? b) ◦ d ,
(21)
by the monotonicity of seminorm. From (20) and the property T (x, 1) = x we see that
(a ? b) ◦ c > (a ◦ c) ? b = (a ◦ c) ? (b ◦ 1) > (a ◦ c) ? (b ◦ d)
(22)
(a ? b) ◦ d > a ? (b ◦ d) = (a ◦ 1) ? (b ◦ d) > (a ◦ c) ? (b ◦ d).
(23)
and
Combining (21) with (22) and (23) yields (17), i.e.,
(a ? b) ◦ (c ∧ d) > (a ◦ c) ? (b ◦ d) for all a, b, c, d ∈ [0, 1]. 11
Now we consider the case of M= ∧ and for several choices of operators ?, ♦, ◦i and sets Y provide, by the use of Theorem 2.4, the characterizations of classes of functions φi for which the corresponding inequality (16) is satis ed. 1.
Let Y = [0, y¯], 0 < y¯ 6 ∞, and ♦ = ? = ◦i = ∧ for all i. Then (16) takes the
form
φ−1 1
Z Z Z −1 −1 φ1 (f ∧ g)dµ > φ2 (S) φ2 (f )dµ ∧ φ3 (S) φ3 (g)dµ , (S) B
A
A∩B
(24) where (S)
Z
hdµ = sup y ∧ µ(A ∩ {h > y}) is the classical Sugeno integral. The
A
y∈Y
inequality (24) holds i −1 −1 −1 a ∧ b ∧ φ−1 1 (c) ∧ φ1 (d) > a ∧ φ2 (c) ∧ b ∧ φ3 (d),
a, b, c, d ∈ Y.
(25)
−1 Taking a = b with b = max φ−1 i (c), φi (d) we obtain the following condition i
−1 −1 −1 φ−1 1 (c) ∧ φ1 (d) > φ2 (c) ∧ φ3 (d),
c, d ∈ Y,
which is equivalent to (25). Substituting su ciently large d and c yields φ−1 1 (c) > −1 −1 φ−1 2 (c) and φ1 (d) > φ3 (d). Hence the condition φ1 (c) 6 φi (c), i = 2, 3, is nec-
essary and su cient for the validity of (24). In particular, it holds in the case of
Y = [0, 1], φ1 (x) = x, φ2 (x) = xp and φ3 (x) = xq , where 0 < p, q < 1, corresponding to Hölder's inequality (see [13]). 2.
Let ? = ♦ = · and ◦i = ∧. We will determine all the functions φi for which Z Z Z −1 −1 −1 φ1 (S) φ1 (f g)dµ > φ2 (S) φ2 (f )dµ φ3 (S) φ3 (g)dµ . (26) A∩B
A
B
From Theorem 2.4 it follows that φi have to ful ll the condition
−1 −1 −1 (ab) ∧ φ−1 1 (c) ∧ φ1 (d) > a ∧ φ2 (c) b ∧ φ3 (d) , 12
a, b, c, d ∈ Y.
(27)
We will consider rst the case of Y = [0, 1]. Putting a = b = d = 1 gives φ−1 1 (c) > −1 −1 φ−1 2 (c) for c ∈ Y as φi (Y ) = Y. Taking a = b = c = 1 yields φ1 (d) > φ3 (d) for
d ∈ Y. Hence φ1 (x) 6 φi (x) for i = 2, 3. We claim that the last condition implies (27). To see this note that for any x, y > 0 (28)
1 ∧ (xy) > (1 ∧ x)(1 ∧ y).
Putting x = α/a and y = β/b we obtain that for arbitrary positive a, b and nonnegative α, β (29)
(ab) ∧ (αβ) > (a ∧ α)(b ∧ β). Therefore
−1 −1 −1 (ab) ∧ φ−1 1 (c) ∧ φ1 (d) > (ab) ∧ φ2 (c) ∧ φ3 (d) −1 > (ab) ∧ φ−1 2 (c)φ3 (d) −1 > a ∧ φ−1 (c) b ∧ φ (d) . 2 3 We have thus proved that the condition φ1 (x) 6 φi (x), i = 2, 3, is equivalent to (27). Now we turn to the case of Y = [0, ∞]. Putting a = b = ∞ into (27) we get −1 −1 −1 φ−1 1 (c) ∧ φ1 (d) > φ2 (c)φ3 (d),
c, d ∈ Y.
Since φi (Y ) = Y for i = 1, 2, 3, there are no functions φi satisfying (26) for all comonotone f|A and g|B . 3.
Set Y = [0, 1], ♦ = ? = ◦i = · and φi = φ for all i. We examine for which φ's Z Z Z −1 −1 −1 φ (N ) φ(f g)dµ > φ (N ) φ(f )dµ φ (N ) φ(g)dµ , A∩B
where (N )
Z
A
B
hdµ is the N -integral, called also the Shilkret integral (see [23]). In
A
view of Theorem 2.4 we have to nd all increasing functions φ : Y → Y such that 13
φ(Y ) = Y and φ−1 φ(ab)(c ∧ d) > φ−1 φ(a)c φ−1 φ(b)d ,
a, b, c, d ∈ Y.
(30)
It is easy to check that (30) holds if φ(t) = tp , p > 0. We show that only these functions ful ll (30). Observe that φ(1) = 1. Putting d = 1 into (30) we have
−1 φ(ab)c > φ φ φ(a)c b ,
a, b, c ∈ Y.
Writing x = φ−1 φ(a)c we see that φ(ab) φ(xb) > φ(a) φ(x)
(31)
for a, b, x ∈ Y such that a, x 6= 0. Substituting x = 1 into (31) gives
φ(ab) > φ(a)φ(b),
a, b ∈ [0, 1],
while substituting a = 1 into (31) yields
φ(x)φ(b) > φ(xb),
x, b ∈ [0, 1].
We thus get φ(x)φ(y) = φ(xy), x, y ∈ [0, 1]. The continuity of φ and the condition
φ(1) = 1 implies that φ(x) = xp , p > 0 (see, e.g., [9]). 4.
Assume that Y = [0, 1], ♦ = ? = ◦i = · and φ1 (x) = x. We will determine all the
functions φ2 and φ3 such that the following Hölder type inequality for N -integral Z Z Z −1 −1 f gdµ > φ2 (N ) φ2 (f )dµ φ3 (N ) φ3 (g)dµ , (N ) (32) A∩B
A
B
holds true. The necessary and su cient condition for (32) is of the form
ab(c ∧ d) > φ−1 φ2 (a)c φ−1 φ3 (b)d . 2 3
(33)
Putting b = d = 1 gives
φ2 (ac) > φ2 (a)c, 14
a, c ∈ Y,
(34)
while taking a = c = 1 leads to φ3 (bd) > φ3 (b)d, where b, d ∈ Y. Hence if the functions φi satisfy (33), then
φi (ac) φi (a) > , ac a
0 < a, c 6 1.
(35)
The inequality (35) means that φi (x)/x are nonincreasing on (0, 1]. The conditions (33) and (35) are equivalent as
−1 φ−1 φ (a)c φ φ (b)d 6 (ac)(bd) 6 ab(c ∧ d) 2 3 2 3 for a, b, c, d ∈ Y. We thus have proved that (33) is ful lled for a pair φ2 , φ3 i
φi (x)/x are nonincreasing. Examples of such functions include φi (x) = xpi with 0 < pi < 1 as well as φi (x) = 1.5x1[0,1/2] (x) + 0.5(x + 1)1[1/2,1] (x), in which i = 2, 3. 5.
The reverse Minkowski inequality Z Z Z −1 −1 −1 φ φ(f + g) ◦ µ > φ φ(f ) ◦ µ + φ φ(g) ◦ µ
is not valid for the Sugeno integral (◦ = ∧), the N -integral (◦ = ·) and for any other integrals with respect to the operators ◦ such that a ◦ 0 = 0 and a ◦ 1 > 0 for a > 0. To see this observe that putting c = 0 into the inequality −1
φ
−1 −1 φ(a + b) ◦ (c ∧ d) > φ φ(a) ◦ c + φ φ(b) ◦ d
we get the condition 0 > φ(b) ◦ d for b, d ∈ Y, which is not satis ed for any increasing function provided 1 ∈ Y. The above examples concern the case of comonotone functions f and g corresponding to M= ∧, which is the only one which has been studied previously. Now 15
we will consider the case when functions f and g are positively dependent with respect to the product operator. 6.
Let M= ·, ? = ♦ = · and ◦i = ∧. Our goal is to determine functions φi for
which
φ−1 1
Z Z Z −1 −1 φ1 (f g)dµ > φ2 (S) φ2 (f )dµ φ3 (S) φ3 (g)dµ , (S) B
A
A∩B
(36)
where f|A and g|B are positively dependent with respect to ·. To do that we only need to nd the functions φi ful lling the inequality
−1 −1 (ab) ∧ φ−1 (cd) > a ∧ φ (c) b ∧ φ (d) , 1 2 3
(37)
where a, b, c, d ∈ Y = [0, 1]. Writing a = b = 1 we have −1 −1 φ−1 1 (cd) > φ2 (c)φ3 (d),
c, d ∈ [0, 1].
(38)
We will show the equivalency of (37) and (38). If (38) holds, then
−1 −1 −1 −1 (ab) ∧ φ−1 1 (cd) > (ab) ∧ φ2 (c)φ3 (d) > a ∧ φ2 (c) b ∧ φ3 (d) , by (29), so (37) is satis ed. Substituting c = φ2 (x) and d = φ3 (y) into (38) leads to the equivalent condition:
φ2 (x)φ3 (y) > φ1 (xy),
x, y ∈ [0, 1].
(39)
Thus we have showed that (36) is valid for all positively dependent functions f|A and g|B i the functions φi ful ll the condition (39). For example, φi (x) = xp , where
p > 0, i = 1, 2, 3, as well as φ1 (x) = x and arbitrary increasing functions φ2 and φ3 such that φi (x)/x is nonincreasing and φi [0, 1] = [0, 1], meet the condition (39). The latter statement is a consequence of the fact that φ2 (x)/x φ3 (y)/y > φ2 (1)φ3 (1) = 1. 16
7.
Let M be the ukasiewicz norm on [0, 1] (see Example 2.3), ◦i = · for all i and
? = ♦. In the case of φi (x) = x, i = 1, 2, 3, the necessary and su cient condition to the validity of (4) for positively dependent with respect to M functions f and g is
(a ? b)(c + d − 1)+ > (ac) ? (bd),
a, b, c, d ∈ [0, 1].
(40)
We show that the continuous operator a ? b = (a + b − 1)+ ful lls (40). Obviously, (40) holds true if ac + bd < 1. Suppose ac + bd ≥ 1. Since a + b > ac + bd > 1 and c + d > ac + bd > 1, we see that for ac + bd > 1, (40) takes the form
(a + b − 1)(c + d − 1) > ac + bd − 1. The last inequality is true because it is equivalent to the following one: (a − 1)(d − 1) + (b − 1)(c − 1) > 0. An interesting open problem is to characterize the class of left-continuous operators ? satisfying the condition (40).
3
Related results
Throughout this section we shall use the notation and work under the assumptions introduced before the statement of Theorem 2.1. The following theorem gives the su cient conditions to the validity of the reverse Chebyshev-type inequality, i.e., the inequality opposite to (4). Theorem 3.1.
Assume Y is monotonically closed, b 7→ a ◦1 b is nondecreasing for
a ∈ Y, ? is an increasing operator and Y ? Y = Y, where Y ? Y = {a ? b : a, b ∈ Y } .
Assume also that (B1) for a xed A ∈ F and for any a, b ∈ Y and M : µ(F) × µ(F) → µ(F) µ
f|A > a ∪ g|A > b 6 µ f|A > a M µ g|A > b ,
17
(41)
(B2) for any a, b ∈ Y and c, d ∈ µ(F) φ−1 φ1 (a ? b) ◦1 (c M d) 6 φ−1 φ2 (a) ◦2 c ♦ φ−1 φ3 (b) ◦3 d . 1 2 3
Then φ−1 1
Z
φ1 (f ? g) ◦1 µ
A
Proof.
6
φ−1 2
Z
φ2 (f ) ◦2 µ
A
♦ φ−1 3
Z
φ3 (g) ◦3 µ .
(42)
A
Since ? is increasing, we have that f (x) ? g(x) < a ? b if f (x) < a and
g(x) < b. Hence for any A A ∩ {f ? g > a ? b} ⊂ A ∩ {f > a} ∪ {g > b} = A ∩ {f > a} ∪ A ∩ {g > b} . (43) From B1 we conclude that for a, b ∈ Y φ−1 φ (a ? b) ◦ µ A ∩ {φ (f ? g) > φ (a ? b)} 1 1 1 1 1 6 φ−1 φ (a ? b) ◦ µ(A ∩ {φ (f ) > φ (a)}) M µ(A ∩ {φ (g) > φ (b)}) . 1 1 2 2 3 3 1 From B2 we obtain −1 φ1 φ1 (a ? b) ◦1 µ A ∩ {φ1 (f ? g) > φ1 (a ? b)} −1 −1 6 φ2 φ2 (a) ◦2 µ {A ∩ φ2 (f ) > φ2 (a)} ♦ φ3 φ3 (b) ◦3 µ {A ∩ φ3 (g) > φ3 (b)} . (44) Taking the supremums with respect to a ∈ Y and b ∈ Y of the right-hand side of (44) rst and of the left-hand side of (44) afterwards completes the proof.
Example 3.1.
Let M= ∨, where a ∨ b = max(a, b), and let Y = [0, 1]. If f and g
are comonotone, then B1 is satis ed for any A ∈ F, and the inequality Z Z Z −1 −1 −1 φ1 (N ) φ1 (f g)dµ 6 φ2 (N ) φ2 (f )dµ ∨ φ3 (N ) φ3 (g)dµ , (45) A
A
A
18
holds true provided that
φ−1 φ1 (ab)(c ∨ d) 6 φ−1 φ2 (a)c ∨ φ−1 φ3 (b)d 1 2 3
(46)
φ3 (b)c > φ2 (ab)c and φ−1 φ2 (a)c > φ−1 for all a, b, c, d ∈ [0, 1]. Since φ−1 3 2 2 φ3 (ab)d , the condition (46) is ful lled if φ−1 3 φi (x)y , φ−1 φ1 (x)y 6 φ−1 1 i
x, y ∈ [0, 1], i = 2, 3.
For example, (45) is valid for φi (x) = φ(x), i = 1, 2, 3, where φ is an increasing function, and for φi (x) = xpi , where 0 < p1 6 p2 6 p3 . It is worth to note that B1 holds for all functions f, g if M= ∨ and µ is the maxitive measure, i.e., µ(A ∪ B) = max µ(A), µ(B) for all A, B. An example of such a measure is the possibility measure µ(A) = sup m(x), where m : X → [0, ∞] is an arbitrary function such that µ(X) > 0.
x∈A
Wu et al. [28] proved that if a ? b > a ∨ b for a, b ∈ [0, ∞] and φ1 (x) > φi (x),
i = 2, 3, then for any comonotone functions f, g Z Z Z −1 −1 −1 φ1 (S) φ1 (f ? g)dµ 6 φ2 (S) φ2 (f )dµ ? φ3 (S) φ3 (g)dµ . (47) A
A
A
Our su cient condition for (47) is of the form
φ−1 φ1 (a ? b) ∧ (c ∨ d) 6 φ−1 φ2 (a) ∧ c ? φ−1 φ3 (b) ∧ d . 1 2 3
(48)
Note that none of the conditions a ? b > a ∨ b and (48) does not imply the other. By the monotonicity of the Sugeno integral, we have Z Z Z (S) f ∧ gdµ 6 (S) f dµ ∧ (S) gdµ A
A
(49)
A
for any f, g and A ∈ F (see [26]). Unfortunately, (49) does not follow neither from Theorem 3.1 and nor from [28] since the operator ? = ∧ is not bounded from below 19
by the maximum. Moreover, the inequality (49) does not follow from Theorem 3.1, even if f and g are comonotone because the condition B2 of the form
(a ∧ b) ∧ (c ∨ d) 6 (a ∧ c) ∧ (b ∧ d),
a, b, c, d > 0
is not ful lled for all a > b > c > d. This means that even in the case of the classical Sugeno integral, the condition B2 is not necessary for (42). Next theorem provides, under some additional restrictions on the operators M and ◦i , the necessary conditions to the validity of (42). Theorem 3.2.
Let Y = [0, y¯] for 0 < y¯ 6 ∞. Suppose that a M 0 > a, 0 M
a > a and a M a > a for a ∈ Y. Suppose also that the functions a 7→ a ◦i b are
nondecreasing and a ◦i 0 = 0 for a ∈ Y, i = 1, 2, 3. If (42) is valid for f, g satisfying (41),
then
(B3) for a, b ∈ Y and c ∈ µ(F) φ−1 φ1 (a ? b) ◦1 c 6 φ−1 φ2 (a) ◦2 c ♦ φ−1 φ3 (b) ◦3 c . 1 2 3
Proof.
Let α, β ∈ Y and A be an arbitrary measurable set. Set f (x) = α and
g(x) = β for all x ∈ X. It is easy to verify that f|A and g|A ful ll the condition B1, Z Z Z φ1 (f ? g) ◦1 µ = φ1 (α ? β) ◦1 µ(A), φ2 (f ) ◦2 µ = φ2 (α) ◦2 µ(A), φ3 (g) ◦3 µ = φ3 (β) ◦3 µ(A). A
A
A
Hence B3 follows from (42).
Example 3.2.
Let M= ∨. It is not true that Z 1 Z 1 Z (S) f gdµ 6 (S) f dµ (S) 0
0
1
gdµ
0
as the inequality B3 of the form (ab)∧c 6 (a∧c)(b∧c) does not hold for a = b = 1. 20
The operators M= ∨ and M= + meet the assumptions of Theorem 3.2 for any
Y, the operator a M b = 1 − (1 − a)(1 − b) meets these assumptions for Y = [0, 1], while the operator M= · does not meet them. We do not know whether there are operators M, ?, ♦, ◦i for which the condition B3 is necessary and su cient. As an open problem we propose to establish the necessary and su cient conditions under which the inequality (42) holds for all functions satisfying (41).
4
Conclusions
We have presented the necessary and su cient conditions to the validity of Chebyshev type and reverse Chebyshev type inequalities for generalized Sugeno integrals in the case of functions belonging to an essentially wider class than the comonotone functions. For speci c choices of operators, we have characterized classes of functions for which the Chebyshev type inequality for the classical Sugeno integral is satis ed. The proofs consist in solving some functional inequalities.
5
Acknowledgments
The authors would like to thank the referees for their valuable comments which led to improvements in the paper. The work on this paper was partially supported by the grant for young researchers from Lodz University of Technology I-2 501 plan 2 number 12/2013.
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MAREK KALUSZKA INSTITUTE OF MATHEMATICS LODZ UNIVERSITY OF TECHNOLOGY UL. WÓLCZASKA 215 90-924 LODZ POLAND E-MAIL: [email protected] ANDRZEJ OKOLEWSKI INSTITUTE OF MATHEMATICS LODZ UNIVERSITY OF TECHNOLOGY UL. WÓLCZASKA 215 24
90-924 LODZ POLAND E-MAIL: [email protected] MICHAL BOCZEK INSTITUTE OF MATHEMATICS LODZ UNIVERSITY OF TECHNOLOGY UL. WÓLCZASKA 215 90-924 LODZ POLAND E-MAIL: [email protected]
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