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Convexity conditions on t-norms and their additive generators
Fuzzy Sets and Systems 151 (2005) 353 – 361 www.elsevier.com/locate/fss
Convexity conditions on t-norms and their additive generators夡 Roberto Ghiselli Riccia,∗ , Mirko Navarab a D.E.I.R., Università di Sassari, 07100 Sassari, Italy b Center for Machine Perception, Department of Cybernetics, Faculty of Electrical Engineering, Czech Technical University,
Technická 2, 166 27 Praha, Czech Republic Received 23 September 2003; received in revised form 7 May 2004; accepted 7 May 2004 Available online 7 June 2004
Abstract Let T be an Archimedean (continuous) t-norm and ∈]0, 21 [. Fodor and Ovchinnikov have studied a relation between the inequality T (max(x − , 0), min(x + , 1)) T (x, x) and the convexity of the additive generator of T . Here we clarify this relation and answer open problems from previous studies. © 2004 Elsevier B.V. All rights reserved. MSC: primary 03E72; secondary 20M05; 26E50; 68T37 Keywords: t-norm; Additive generator; Convexity
1. Motivation and formulation of the problems The class of continuous Archimedean t-norms is very large. We sometimes simplify our study by looking at their restrictions. When we restrict a t-norm T to the diagonal D = {(y, y) : y ∈ [0, 1]}, we obtain the diagonal section, T (y) = T (y, y). Diagonal sections of continuous Archimedean t-norms were fully characterized in [10]. It is also known that a continuous Archimedean t-norm is not uniquely determined by its diagonal section [4]; all continuous Archimedean t-norms with a given diagonal section were characterized in [10]. We may also consider a restriction to the other diagonal of the
夡
This research was supported by grant 201/02/1540 of the Grant Agency of the Czech Republic.
∗ Corresponding author. Tel.: +39-333-6659-159; fax: +39-0789-646108.
E-mail addresses: [email protected] (R.G. Ricci), [email protected] (M. Navara). 0165-0114/$ - see front matter © 2004 Elsevier B.V. All rights reserved. doi:10.1016/j.fss.2004.05.005
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unit square, D1 = {(y, z) : y, z ∈ [0, 1], y + z = 1} = {(y, 1 − y) : y ∈ [0, 1]}. According to [1] (see also a recent overview in Section 7.2 of [5]), each strict t-norm is fully determined by its restriction to D ∪ D1 , or, more generally, to D ∪ D for some ∈]0, 2[, where D = {(y, z) : y, z ∈ [0, 1], y + z = } = {(y, − y) : y ∈ [max( − 1, 0), min(, 1)]}. Fodor and Ovchinnikov have noticed a convexity property of the restriction T D which is related to the convexity of the additive generator t of T. They studied the inequality T (y, z) + T (z, y) T (y, z) = 2 y+z y+z T , , y, z ∈ [0, 1] 2 2 which, with the substitutions y := x − ,
z := x +
(1)
becomes equivalent to T (x − , x + ) T (x, x).
(2)
Fodor and Ovchinnikov [2] posed the following question (Open Problem 5 in [6]): Problem 1 (Weak version). Let T be a continuous Archimedean t-norm with additive generator t: [0, 1] → [0, ∞] and ∈]0, 21 [. Prove or disprove that T (max(x − , 0), min(x + , 1)) T (x, x)
(3)
holds for all x ∈ [0, 1] if and only if t is convex. A stronger version of the problem is formulated as Problem 5 in [7]: Problem 2 (Strong version). Let T be a continuous Archimedean t-norm with additive generator t: [0, 1] → [0, ∞]. Prove or disprove that (3) holds for all ∈]0, 21 [ and x ∈ [0, 1] if and only if t is convex. Note that a positive solution to this problem would induce a new characterization of associative copulas [11]. However, we give negative answers to both problems in the sequel. Remark 1. Notice that (3) is trivially satisfied for x , because then it attains the form 0 T (x, x). Thus we shall consider (3) only for x > .
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355
As another motivation, for most examples of strict t-norms found in the literature, the maximum on is attained at (x, x), D2x T (x, x) = max T. D2x
(4)
This condition is equivalent to the satisfaction of (2) for all ∈ [0, min(x, 1 − x)]. Equality (4) is not satisfied in general (it does not follow from the monotonicity of the t-norm). Counterexamples can be found only among t-norms with nonconvex additive generators. Thus convexity of the additive generator is sufficient for (3) being satisfied for all ∈]0, 21 [ and x ∈ [0, 1]. The question is whether it is also necessary. Condition (4) plays also an important role in the T-sum of fuzzy numbers [8]. Normally, this operation requires two nested cycles to cover the space of two variables. If (4) is satisfied for all x, instead of the t-norm T we need only its diagonal section and a single cycle over one variable suffices. This reduces the order of complexity. Further reduction is possible if the T-sum preserves the shape of fuzzy numbers; then only the spread of the resulting fuzzy number is computed. This direction of research has been outlined in [9] and condition (4) is relevant also in this context. Remark 2. Convexity of the multiplicative generator is not related to the convexity of the additive generator which is studied here.
2. Strict t-norms In this section, we suppose that T is a strict t-norm and t: [0, 1] → [0, ∞] is its additive generator, i.e., a strictly decreasing bijection such that T (y, z) = t −1 (t (y) + t (z)) for all y, z ∈ [0, 1]. Remark 3. In view of Remark 1, we consider inequality (3) for ∈ ]0, 21 [ and x ∈], 1]. We distinguish two cases: (A) when < x < 1 − , then (3) is equivalent to t (x − ) + t (x + ) 2 t (x);
(5)
(B) when x 1 − , then (3) is equivalent to t (x − ) 2 t (x).
(6)
In the sequel, we shall use the following two lemmas, showing the proof only of the second, because the first is a classical result on convexity (see, e.g., [3]).
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Lemma 1. A continuous function t: [0, 1] → R is convex iff y+z t (y) + t (z) 2 t 2
(7)
for all y, z ∈ [0, 1]. Lemma 2. A continuous function t: [0, 1] → R is convex iff t (x − ) + t (x + ) 2 t (x)
(8)
for all ∈]0, 21 [ and x ∈], 1 − [. Proof. For (x − , x + ) in the interior of [0, 1]2 , we apply Lemma 1 and the fact that (8) is equivalent to (7) with substitutions (1). The remaining case when (x − , x + ) is on the boundary of [0, 1]2 follows by continuity. First, let us verify that the convexity of the additive generator is sufficient for (3). Proposition 1. Let T be a strict t-norm with a convex additive generator. Then (3) is satisfied for all ∈]0, 21 [ and x ∈ [0, 1]. Proof. According to Remark 3, we can separately analyse two cases: when < x < 1 − , then (3) is equivalent to (5) which is just a special form of convexity. If x 1 − , then (3) is equivalent to (6) which follows from 0
t (x − ) t (2x − 1) = t (2x − 1) + t (1) 2 t (x), where the last inequality is due to convexity of t in the form (7).
Note that strict t-norms with a convex additive generator are exactly strict copulas. Proposition 1 and the following give a positive answer to Problem 2 for strict t-norms: Proposition 2. Let T be a strict t-norm such that (3) is satisfied for all ∈]0, 21 [ and x ∈ [0, 1]. Then the additive generator of T is convex. Proof. According to Lemma 2, it suffices to verify convexity in the form (8). The corresponding values x and satisfy the assumption < x < 1 − of case (A) of Remark 3, hence convexity of t is proved. Problem 1 asks whether Proposition 2 can be strengthened so that it is sufficient to assume (3) for a single value ∈]0, 21 [. We show that this is not the case. Example 1. Given any ∈]0, 21 [, there exists a strict t-norm T with a nonconvex additive generator such that (3) holds for all x ∈ [0, 1].
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to
357
For a fixed ∈]0, 21 [, with the substitution x := ε + , ε ∈]0, 1−2[, inequality (5) becomes equivalent t (ε) + t (ε + 2) 2 t (ε + ).
(9)
The left-hand side, considered as a function (ε) = t (ε)+t (ε +2), is continuous and strictly decreasing, with limε→0+ (ε) = +∞, so there exists ε0 ∈]0, 1 − 2[ such that (ε0 ) 2 t (). For all ε ∈]0, ε0 [, we obtain t (ε) + t (ε + 2) = (ε) > (ε0 ) 2 t () > 2 t (ε + ). Thus (9), and hence also (5), is satisfied independently of the choice of t ]0, ε0 [. If we take for t a decreasing bijection which is convex on [ε0 , 1], then (3) is satisfied for all x ∈ [0, 1]. Obviously, t may be chosen nonconvex on ]0, ε0 [. A counterexample to Problem 1 may be constructed also on a completely different principle as follows. For ∈]0, 21 [, let a continuous function t on [0, 1] be called -convex if (7) holds for all y, z ∈ [0, 1] with |y − z| = 2. Lemma 2 says that t is convex iff it is -convex for every ∈]0, 21 [. Fundamentally, on the basis of this definition for one fixed value , the -convexity links points of distance 2 (and the middle of the interval with the distance from the bounds), but it does not impose any relation between the values on closer points. Based on the above observation, we may construct another counterexample to the original problem: Example 2. Let t (y) = − ln y +
1 (1 − cos 6y). 6
For y < 1, we have 1 t (y) = − + sin 6y < 0, y 1 <−1
so t is strictly decreasing and it is an additive generator of a t-norm T. The second derivative, t (y) =
1 + 6 cos 6y, y2
attains negative values, e.g., 36 5 = − 6 < 0, t 6 25 hence t is not convex. Nevertheless, we shall prove that the corresponding t-norm T satisfies (3) for = 13 . We distinguish two cases from Remark 3.
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For x ∈
1 2 3, 3
, inequality (3) attains the form (5), explicitly 1 1 2 − ln x − − ln x + + (1 − cos 6x) 3 3 6 1 2 − ln x + (1 − cos 6x) 6
which is equivalent to
− ln x − 13 − ln x + 13 − 2 ln x
and valid due to the convexity of the function − ln. For x ∈ 23 , 1 , inequality (3) attains the form (6), explicitly 1 1 1 − ln x − + (1 − cos 6x) 2 − ln x + (1 − cos 6x) . 3 6 6 This is equivalent to ln
x2 x−
1 3
1 (1 − cos 6x) 6
and valid, because the left-hand side is increasing on ln
2 2
x2 x−
1 3
4
1
2
3, 1
and
1
ln 2 3 1 = ln > (1 − cos 6x). 3 3 6 3 − 3
Thus (3) is satisfied.
1 , but not for all e.g., x = 21 . Example 2 shows that (4) may be valid for some values of x e.g., x = 12 We may extend the task by considering more (but not all) values of : Problem 3 (Generalized version). Let A ⊆]0, 21 [ and let T be a continuous Archimedean t-norm with additive generator t: [0, 1] → [0, ∞]. Prove or disprove that (3) holds for all ∈ A and x ∈ [0, 1] if and only if t is convex. Problem 3 is a common generalization of Problem 2 (with A =]0, 21 [ ) and Problem 1 (with A a singleton). We answer Problem 3 for strict t-norms in the following theorem, thus generalizing all previous results. Theorem 1. Let T be a strict t-norm with additive generator t: [0, 1] → [0, ∞]. For A ⊆]0, 21 [, the condition inf A = 0 is necessary and sufficient for the equivalence between the convexity of t and the satisfaction of (3) for all ∈ A and x ∈ [0, 1]. Proof. According to Proposition 1, the convexity of t implies (3) for all ∈]0, 21 [ and x ∈ [0, 1]. It remains to prove that the reverse implication holds iff inf A = 0.
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If inf A > 0, the technique of Example 1 (applied to inf A in place of ) allows us to construct a nonconvex additive generator t of a strict t-norm T satisfying (3) for all ∈ A and x ∈ [0, 1]. Finally, suppose that inf A = 0. We shall prove the convexity of t by contradiction. Using Lemma 1, we assume that t violates convexity in the form (7). Thus there are y, z ∈ [0, 1] such that y+z t (y) + t (z) < 2 t . (10) 2 Due to the continuity of t, inequality (10) holds also in some neighbourhoods of y, z. We may find a sufficiently small ∈ A and y , z ∈ [0, 1] (in the neighbourhoods of y, z, respectively) satisfying (10) and such that |z − y |/2 is an integer multiple of . This contradicts the -convexity of t, because the -convexity implies the reverse inequality in (10) for all y , z ∈ [0, 1] such that |z − y |/2 is an integer multiple of . The proof is complete.
3. Nilpotent t-norms The original open problem was formulated for continuous Archimedean t-norms. Having solved it for strict t-norms, we now concentrate on the remaining case of nilpotent ones. In this section, we suppose that T is a nilpotent t-norm and t: [0, 1] → [0, ∞[ is its additive generator, i.e., a strictly decreasing continuous function such that t (1) = 0 and T (y, z) = t −1 (min (t (y) + t (z), t (0))) for all y, z ∈ [0, 1]. As the additive generator of a nilpotent t-norm is unique up to a positive multiple, we assume here, without loss of generality, that t is normalized so that t (0) = 1. We use the notation = t −1 21 ∈]0, 1[. (Without normalization, is the unique value satisfying t () = 21 t (0). In terms of the t-norm T, it can be expressed as = max{x ∈ [0, 1] : T (x, x) = 0} = max −1 T ({0}).) Remark 4. As we shall see later (in the proof of Theorem 2), the case > 21 contradicts (3), therefore we shall restrict to the case ∈]0, 21 ]. In view of Remark 1, we consider inequality (3) for ∈]0, 21 [, x ∈], 1], and we distinguish three cases: (A) when < x < 1 − and x > , then (3) is equivalent to (5); (B) when x 1 − , then (3) is equivalent to (6); (C) when < x , then (3) is equivalent to t (x − ) + t (x + ) 1. The situation for nilpotent t-norms appears to be quite different from the case of strict t-norms, because even Problem 2 has a negative answer (which implies negative answers to Problems 1 and 3): Example 3. There exists a nilpotent t-norm T with a nonconvex additive generator such that (3) holds for all ∈]0, 21 [ and x ∈ [0, 1]. Let us take the nilpotent t-norm T with additive generator t (y) =
1 2
(1 + cos y).
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R.G. Ricci, M. Navara / Fuzzy Sets and Systems 151 (2005) 353 – 361
Here = 21 . The left-hand side of (5) is t (x − ) + t (x + ) = 1 + 21 (cos (x − ) + cos (x + )) = 1 + cos cos x. For 21 < x < 1 − , case (A) occurs and (3) is satisfied, as it is easy to verify, because cos x < 0. For x 1 − , case (B) occurs and (3) follows from the convexity of t on 21 , 1 . For < x 21 , we obtain 2 t (x) 1, T (x, x) = 0, and also t (x − ) + t (x + ) = 1 + cos · cos x 1, as cos x 0, hence T (max(x − , 0), min(x + , 1)) = 0.
We proved that T satisfies (3), although its additive generator is not convex on 0, 21 . The following result gives a complete characterization of nilpotent t-norms which satisfy (3) for all ∈]0, 21 [ and x ∈ [0, 1].
Theorem 2. Let T be a nilpotent t-norm with additive generator t. Denote = t −1 conjunction of the two conditions
1 2
t (0) . The
(C1) for all ∈]0, min(, 1 − )] we have t ( − ) + t ( + ) 1, (C2) t is convex on [, 1], is necessary and sufficient for T to verify (3) for all ∈]0, 21 [ and x ∈ [0, 1]. Proof. Without loss of generality, we assume a normalized additive generator, so t (0) = 1 and t () = 21 . Suppose first that T satisfies (3) for all ∈]0, 21 [ and x ∈ [0, 1]. As T (, ) = 0, inequality (3) implies T ( − , + ) = 0 for all ∈]0, min(, 1 − )] and this is equivalent to (C1). Notice that condition (C1) implies 21 . Indeed, if > 21 , then substitution := 1 − in (C1) gives t (2 − 1) 1 = t (0) which contradicts the strict monotonicity of t. Thus (C1) applies to ∈]0, ]. To prove (C2), it suffices to verify inequality (7) for t and for y, z ∈ [, 1]. This follows from (5) after substitutions (1) (for x and satisfying the assumptions of case (A)). To prove the reverse implication, assume that t: [0, 1] → [0, 1] is a decreasing bijection satisfying
(C1), (C2). Hence 21 . To verify (3) for the corresponding t-norm T, we shall distinguish cases from Remark 4. Case (C) follows from (C1): t (x − ) + t (x + ) t ( − ) + t ( + ) 1. In case (A), we have to prove the -convexity condition (5). For x − , it follows from the convexity of t on [, 1]. If x − < < x, then (5) still holds. To prove this, let sy,z denote the linear function (secant of the graph of t) determined by the points (y, t (y)), (z, t (z)). We apply (C1) to x − in place of − (and 2 − x + in place of + ) and obtain that sx−,2−x+ lies above the graph of t on [, 2 − x + ] and below the graph of t on [2 − x + , 1]. (Here we used the convexity of t on [, 1]; by “above’’ and “below’’ we mean nonstrict inequalities.) As x > , we have x + ∈ [2 − x + , 1] and hence also
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361
sx−,x+ () sx−,2−x+ () t (). As sx−,x+ (x + ) = t (x + ), the convexity of t on [, 1] implies sx−,x+ (x) t (x) which is equivalent to (5). Case (B) is solved analogously to case (A), (6) is a weaker condition than convexity. Remark 5. Condition (C1) can be equivalently formulated as follows: 21
and
t (y) 1 − t (2 − y) for all y ∈ [0, ].
Remark 6. During the reviewing process, we have been informed of a result by Marková [8]. Although it dealt with a differently formulated problem (the shape of the T-sum of L-R-fuzzy numbers), it induces a solution of Problem 2 and in the case of triangular fuzzy numbers it leads to the same characterization as in Propositions 1–2 and Theorem 2. On the other hand, it gives another motivation to these problems. Our work extends the results of [8] by considering three different versions of the original problem posed by Fodor and Ovchinnikov, proving Theorem 1 and presenting examples that limit further strengthening. We also contribute by the study of -convexity as a weaker form of convexity. Acknowledgements The authors would like to thank R. Mesiar for his valuable comments and inspiration. References [1] J.P. Bézivin, M.S. Tomás, On the determination of strict t-norms on some diagonal segments, Aequations Math. 45 (1993) 239–245. [2] J. Fodor, S. Ovchinnikov, personal communication, 2003. [3] G.H. Hardy, J.E. Littlewood, G. Pólya, Inequalities, 2nd Edition, Cambridge University Press, Cambridge, 1952. [4] C. Kimberling, On a class of associative functions, Publ. Math. Debrecen 20 (1973) 21–39. [5] E.P. Klement, R. Mesiar, E. Pap, Triangular Norms, Kluwer, Dordrecht, Boston, London, 2000. [6] E.P. Klement, R. Mesiar, E. Pap, Triangular norms: some open questions, Proc. Linz Seminar 2003, Johannes Kepler University Linz, 2003, pp. 135–138. [7] E.P. Klement, R. Mesiar, E. Pap, Problems on triangular norms and related operators, Fuzzy Sets and Systems 145 (2004) 471–479. [8] A. Marková, T-sum of L-R-fuzzy numbers, Fuzzy Sets and Systems 85 (1997) 379–384. [9] R. Mesiar, Shape preserving additions of fuzzy intervals, Fuzzy Sets and Systems 86 (1997) 73–78. [10] R. Mesiar, M. Navara, Diagonals of continuous triangular norms, Fuzzy Sets and Systems 104 (1999) 34–41. [11] R.B. Nelsen, An Introduction to Copulas, Springer, Berlin, 1999.
Convexity conditions on t-norms and their additive generators夡 Roberto Ghiselli Riccia,∗ , Mirko Navarab a D.E.I.R., Università di Sassari, 07100 Sassari, Italy b Center for Machine Perception, Department of Cybernetics, Faculty of Electrical Engineering, Czech Technical University,
Technická 2, 166 27 Praha, Czech Republic Received 23 September 2003; received in revised form 7 May 2004; accepted 7 May 2004 Available online 7 June 2004
Abstract Let T be an Archimedean (continuous) t-norm and ∈]0, 21 [. Fodor and Ovchinnikov have studied a relation between the inequality T (max(x − , 0), min(x + , 1)) T (x, x) and the convexity of the additive generator of T . Here we clarify this relation and answer open problems from previous studies. © 2004 Elsevier B.V. All rights reserved. MSC: primary 03E72; secondary 20M05; 26E50; 68T37 Keywords: t-norm; Additive generator; Convexity
1. Motivation and formulation of the problems The class of continuous Archimedean t-norms is very large. We sometimes simplify our study by looking at their restrictions. When we restrict a t-norm T to the diagonal D = {(y, y) : y ∈ [0, 1]}, we obtain the diagonal section, T (y) = T (y, y). Diagonal sections of continuous Archimedean t-norms were fully characterized in [10]. It is also known that a continuous Archimedean t-norm is not uniquely determined by its diagonal section [4]; all continuous Archimedean t-norms with a given diagonal section were characterized in [10]. We may also consider a restriction to the other diagonal of the
夡
This research was supported by grant 201/02/1540 of the Grant Agency of the Czech Republic.
∗ Corresponding author. Tel.: +39-333-6659-159; fax: +39-0789-646108.
E-mail addresses: [email protected] (R.G. Ricci), [email protected] (M. Navara). 0165-0114/$ - see front matter © 2004 Elsevier B.V. All rights reserved. doi:10.1016/j.fss.2004.05.005
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R.G. Ricci, M. Navara / Fuzzy Sets and Systems 151 (2005) 353 – 361
unit square, D1 = {(y, z) : y, z ∈ [0, 1], y + z = 1} = {(y, 1 − y) : y ∈ [0, 1]}. According to [1] (see also a recent overview in Section 7.2 of [5]), each strict t-norm is fully determined by its restriction to D ∪ D1 , or, more generally, to D ∪ D for some ∈]0, 2[, where D = {(y, z) : y, z ∈ [0, 1], y + z = } = {(y, − y) : y ∈ [max( − 1, 0), min(, 1)]}. Fodor and Ovchinnikov have noticed a convexity property of the restriction T D which is related to the convexity of the additive generator t of T. They studied the inequality T (y, z) + T (z, y) T (y, z) = 2 y+z y+z T , , y, z ∈ [0, 1] 2 2 which, with the substitutions y := x − ,
z := x +
(1)
becomes equivalent to T (x − , x + ) T (x, x).
(2)
Fodor and Ovchinnikov [2] posed the following question (Open Problem 5 in [6]): Problem 1 (Weak version). Let T be a continuous Archimedean t-norm with additive generator t: [0, 1] → [0, ∞] and ∈]0, 21 [. Prove or disprove that T (max(x − , 0), min(x + , 1)) T (x, x)
(3)
holds for all x ∈ [0, 1] if and only if t is convex. A stronger version of the problem is formulated as Problem 5 in [7]: Problem 2 (Strong version). Let T be a continuous Archimedean t-norm with additive generator t: [0, 1] → [0, ∞]. Prove or disprove that (3) holds for all ∈]0, 21 [ and x ∈ [0, 1] if and only if t is convex. Note that a positive solution to this problem would induce a new characterization of associative copulas [11]. However, we give negative answers to both problems in the sequel. Remark 1. Notice that (3) is trivially satisfied for x , because then it attains the form 0 T (x, x). Thus we shall consider (3) only for x > .
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355
As another motivation, for most examples of strict t-norms found in the literature, the maximum on is attained at (x, x), D2x T (x, x) = max T. D2x
(4)
This condition is equivalent to the satisfaction of (2) for all ∈ [0, min(x, 1 − x)]. Equality (4) is not satisfied in general (it does not follow from the monotonicity of the t-norm). Counterexamples can be found only among t-norms with nonconvex additive generators. Thus convexity of the additive generator is sufficient for (3) being satisfied for all ∈]0, 21 [ and x ∈ [0, 1]. The question is whether it is also necessary. Condition (4) plays also an important role in the T-sum of fuzzy numbers [8]. Normally, this operation requires two nested cycles to cover the space of two variables. If (4) is satisfied for all x, instead of the t-norm T we need only its diagonal section and a single cycle over one variable suffices. This reduces the order of complexity. Further reduction is possible if the T-sum preserves the shape of fuzzy numbers; then only the spread of the resulting fuzzy number is computed. This direction of research has been outlined in [9] and condition (4) is relevant also in this context. Remark 2. Convexity of the multiplicative generator is not related to the convexity of the additive generator which is studied here.
2. Strict t-norms In this section, we suppose that T is a strict t-norm and t: [0, 1] → [0, ∞] is its additive generator, i.e., a strictly decreasing bijection such that T (y, z) = t −1 (t (y) + t (z)) for all y, z ∈ [0, 1]. Remark 3. In view of Remark 1, we consider inequality (3) for ∈ ]0, 21 [ and x ∈], 1]. We distinguish two cases: (A) when < x < 1 − , then (3) is equivalent to t (x − ) + t (x + ) 2 t (x);
(5)
(B) when x 1 − , then (3) is equivalent to t (x − ) 2 t (x).
(6)
In the sequel, we shall use the following two lemmas, showing the proof only of the second, because the first is a classical result on convexity (see, e.g., [3]).
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Lemma 1. A continuous function t: [0, 1] → R is convex iff y+z t (y) + t (z) 2 t 2
(7)
for all y, z ∈ [0, 1]. Lemma 2. A continuous function t: [0, 1] → R is convex iff t (x − ) + t (x + ) 2 t (x)
(8)
for all ∈]0, 21 [ and x ∈], 1 − [. Proof. For (x − , x + ) in the interior of [0, 1]2 , we apply Lemma 1 and the fact that (8) is equivalent to (7) with substitutions (1). The remaining case when (x − , x + ) is on the boundary of [0, 1]2 follows by continuity. First, let us verify that the convexity of the additive generator is sufficient for (3). Proposition 1. Let T be a strict t-norm with a convex additive generator. Then (3) is satisfied for all ∈]0, 21 [ and x ∈ [0, 1]. Proof. According to Remark 3, we can separately analyse two cases: when < x < 1 − , then (3) is equivalent to (5) which is just a special form of convexity. If x 1 − , then (3) is equivalent to (6) which follows from 0
t (x − ) t (2x − 1) = t (2x − 1) + t (1) 2 t (x), where the last inequality is due to convexity of t in the form (7).
Note that strict t-norms with a convex additive generator are exactly strict copulas. Proposition 1 and the following give a positive answer to Problem 2 for strict t-norms: Proposition 2. Let T be a strict t-norm such that (3) is satisfied for all ∈]0, 21 [ and x ∈ [0, 1]. Then the additive generator of T is convex. Proof. According to Lemma 2, it suffices to verify convexity in the form (8). The corresponding values x and satisfy the assumption < x < 1 − of case (A) of Remark 3, hence convexity of t is proved. Problem 1 asks whether Proposition 2 can be strengthened so that it is sufficient to assume (3) for a single value ∈]0, 21 [. We show that this is not the case. Example 1. Given any ∈]0, 21 [, there exists a strict t-norm T with a nonconvex additive generator such that (3) holds for all x ∈ [0, 1].
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For a fixed ∈]0, 21 [, with the substitution x := ε + , ε ∈]0, 1−2[, inequality (5) becomes equivalent t (ε) + t (ε + 2) 2 t (ε + ).
(9)
The left-hand side, considered as a function (ε) = t (ε)+t (ε +2), is continuous and strictly decreasing, with limε→0+ (ε) = +∞, so there exists ε0 ∈]0, 1 − 2[ such that (ε0 ) 2 t (). For all ε ∈]0, ε0 [, we obtain t (ε) + t (ε + 2) = (ε) > (ε0 ) 2 t () > 2 t (ε + ). Thus (9), and hence also (5), is satisfied independently of the choice of t ]0, ε0 [. If we take for t a decreasing bijection which is convex on [ε0 , 1], then (3) is satisfied for all x ∈ [0, 1]. Obviously, t may be chosen nonconvex on ]0, ε0 [. A counterexample to Problem 1 may be constructed also on a completely different principle as follows. For ∈]0, 21 [, let a continuous function t on [0, 1] be called -convex if (7) holds for all y, z ∈ [0, 1] with |y − z| = 2. Lemma 2 says that t is convex iff it is -convex for every ∈]0, 21 [. Fundamentally, on the basis of this definition for one fixed value , the -convexity links points of distance 2 (and the middle of the interval with the distance from the bounds), but it does not impose any relation between the values on closer points. Based on the above observation, we may construct another counterexample to the original problem: Example 2. Let t (y) = − ln y +
1 (1 − cos 6y). 6
For y < 1, we have 1 t (y) = − + sin 6y < 0, y 1 <−1
so t is strictly decreasing and it is an additive generator of a t-norm T. The second derivative, t (y) =
1 + 6 cos 6y, y2
attains negative values, e.g., 36 5 = − 6 < 0, t 6 25 hence t is not convex. Nevertheless, we shall prove that the corresponding t-norm T satisfies (3) for = 13 . We distinguish two cases from Remark 3.
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For x ∈
1 2 3, 3
, inequality (3) attains the form (5), explicitly 1 1 2 − ln x − − ln x + + (1 − cos 6x) 3 3 6 1 2 − ln x + (1 − cos 6x) 6
which is equivalent to
− ln x − 13 − ln x + 13 − 2 ln x
and valid due to the convexity of the function − ln. For x ∈ 23 , 1 , inequality (3) attains the form (6), explicitly 1 1 1 − ln x − + (1 − cos 6x) 2 − ln x + (1 − cos 6x) . 3 6 6 This is equivalent to ln
x2 x−
1 3
1 (1 − cos 6x) 6
and valid, because the left-hand side is increasing on ln
2 2
x2 x−
1 3
4
1
2
3, 1
and
1
ln 2 3 1 = ln > (1 − cos 6x). 3 3 6 3 − 3
Thus (3) is satisfied.
1 , but not for all e.g., x = 21 . Example 2 shows that (4) may be valid for some values of x e.g., x = 12 We may extend the task by considering more (but not all) values of : Problem 3 (Generalized version). Let A ⊆]0, 21 [ and let T be a continuous Archimedean t-norm with additive generator t: [0, 1] → [0, ∞]. Prove or disprove that (3) holds for all ∈ A and x ∈ [0, 1] if and only if t is convex. Problem 3 is a common generalization of Problem 2 (with A =]0, 21 [ ) and Problem 1 (with A a singleton). We answer Problem 3 for strict t-norms in the following theorem, thus generalizing all previous results. Theorem 1. Let T be a strict t-norm with additive generator t: [0, 1] → [0, ∞]. For A ⊆]0, 21 [, the condition inf A = 0 is necessary and sufficient for the equivalence between the convexity of t and the satisfaction of (3) for all ∈ A and x ∈ [0, 1]. Proof. According to Proposition 1, the convexity of t implies (3) for all ∈]0, 21 [ and x ∈ [0, 1]. It remains to prove that the reverse implication holds iff inf A = 0.
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If inf A > 0, the technique of Example 1 (applied to inf A in place of ) allows us to construct a nonconvex additive generator t of a strict t-norm T satisfying (3) for all ∈ A and x ∈ [0, 1]. Finally, suppose that inf A = 0. We shall prove the convexity of t by contradiction. Using Lemma 1, we assume that t violates convexity in the form (7). Thus there are y, z ∈ [0, 1] such that y+z t (y) + t (z) < 2 t . (10) 2 Due to the continuity of t, inequality (10) holds also in some neighbourhoods of y, z. We may find a sufficiently small ∈ A and y , z ∈ [0, 1] (in the neighbourhoods of y, z, respectively) satisfying (10) and such that |z − y |/2 is an integer multiple of . This contradicts the -convexity of t, because the -convexity implies the reverse inequality in (10) for all y , z ∈ [0, 1] such that |z − y |/2 is an integer multiple of . The proof is complete.
3. Nilpotent t-norms The original open problem was formulated for continuous Archimedean t-norms. Having solved it for strict t-norms, we now concentrate on the remaining case of nilpotent ones. In this section, we suppose that T is a nilpotent t-norm and t: [0, 1] → [0, ∞[ is its additive generator, i.e., a strictly decreasing continuous function such that t (1) = 0 and T (y, z) = t −1 (min (t (y) + t (z), t (0))) for all y, z ∈ [0, 1]. As the additive generator of a nilpotent t-norm is unique up to a positive multiple, we assume here, without loss of generality, that t is normalized so that t (0) = 1. We use the notation = t −1 21 ∈]0, 1[. (Without normalization, is the unique value satisfying t () = 21 t (0). In terms of the t-norm T, it can be expressed as = max{x ∈ [0, 1] : T (x, x) = 0} = max −1 T ({0}).) Remark 4. As we shall see later (in the proof of Theorem 2), the case > 21 contradicts (3), therefore we shall restrict to the case ∈]0, 21 ]. In view of Remark 1, we consider inequality (3) for ∈]0, 21 [, x ∈], 1], and we distinguish three cases: (A) when < x < 1 − and x > , then (3) is equivalent to (5); (B) when x 1 − , then (3) is equivalent to (6); (C) when < x , then (3) is equivalent to t (x − ) + t (x + ) 1. The situation for nilpotent t-norms appears to be quite different from the case of strict t-norms, because even Problem 2 has a negative answer (which implies negative answers to Problems 1 and 3): Example 3. There exists a nilpotent t-norm T with a nonconvex additive generator such that (3) holds for all ∈]0, 21 [ and x ∈ [0, 1]. Let us take the nilpotent t-norm T with additive generator t (y) =
1 2
(1 + cos y).
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R.G. Ricci, M. Navara / Fuzzy Sets and Systems 151 (2005) 353 – 361
Here = 21 . The left-hand side of (5) is t (x − ) + t (x + ) = 1 + 21 (cos (x − ) + cos (x + )) = 1 + cos cos x. For 21 < x < 1 − , case (A) occurs and (3) is satisfied, as it is easy to verify, because cos x < 0. For x 1 − , case (B) occurs and (3) follows from the convexity of t on 21 , 1 . For < x 21 , we obtain 2 t (x) 1, T (x, x) = 0, and also t (x − ) + t (x + ) = 1 + cos · cos x 1, as cos x 0, hence T (max(x − , 0), min(x + , 1)) = 0.
We proved that T satisfies (3), although its additive generator is not convex on 0, 21 . The following result gives a complete characterization of nilpotent t-norms which satisfy (3) for all ∈]0, 21 [ and x ∈ [0, 1].
Theorem 2. Let T be a nilpotent t-norm with additive generator t. Denote = t −1 conjunction of the two conditions
1 2
t (0) . The
(C1) for all ∈]0, min(, 1 − )] we have t ( − ) + t ( + ) 1, (C2) t is convex on [, 1], is necessary and sufficient for T to verify (3) for all ∈]0, 21 [ and x ∈ [0, 1]. Proof. Without loss of generality, we assume a normalized additive generator, so t (0) = 1 and t () = 21 . Suppose first that T satisfies (3) for all ∈]0, 21 [ and x ∈ [0, 1]. As T (, ) = 0, inequality (3) implies T ( − , + ) = 0 for all ∈]0, min(, 1 − )] and this is equivalent to (C1). Notice that condition (C1) implies 21 . Indeed, if > 21 , then substitution := 1 − in (C1) gives t (2 − 1) 1 = t (0) which contradicts the strict monotonicity of t. Thus (C1) applies to ∈]0, ]. To prove (C2), it suffices to verify inequality (7) for t and for y, z ∈ [, 1]. This follows from (5) after substitutions (1) (for x and satisfying the assumptions of case (A)). To prove the reverse implication, assume that t: [0, 1] → [0, 1] is a decreasing bijection satisfying
(C1), (C2). Hence 21 . To verify (3) for the corresponding t-norm T, we shall distinguish cases from Remark 4. Case (C) follows from (C1): t (x − ) + t (x + ) t ( − ) + t ( + ) 1. In case (A), we have to prove the -convexity condition (5). For x − , it follows from the convexity of t on [, 1]. If x − < < x, then (5) still holds. To prove this, let sy,z denote the linear function (secant of the graph of t) determined by the points (y, t (y)), (z, t (z)). We apply (C1) to x − in place of − (and 2 − x + in place of + ) and obtain that sx−,2−x+ lies above the graph of t on [, 2 − x + ] and below the graph of t on [2 − x + , 1]. (Here we used the convexity of t on [, 1]; by “above’’ and “below’’ we mean nonstrict inequalities.) As x > , we have x + ∈ [2 − x + , 1] and hence also
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361
sx−,x+ () sx−,2−x+ () t (). As sx−,x+ (x + ) = t (x + ), the convexity of t on [, 1] implies sx−,x+ (x) t (x) which is equivalent to (5). Case (B) is solved analogously to case (A), (6) is a weaker condition than convexity. Remark 5. Condition (C1) can be equivalently formulated as follows: 21
and
t (y) 1 − t (2 − y) for all y ∈ [0, ].
Remark 6. During the reviewing process, we have been informed of a result by Marková [8]. Although it dealt with a differently formulated problem (the shape of the T-sum of L-R-fuzzy numbers), it induces a solution of Problem 2 and in the case of triangular fuzzy numbers it leads to the same characterization as in Propositions 1–2 and Theorem 2. On the other hand, it gives another motivation to these problems. Our work extends the results of [8] by considering three different versions of the original problem posed by Fodor and Ovchinnikov, proving Theorem 1 and presenting examples that limit further strengthening. We also contribute by the study of -convexity as a weaker form of convexity. Acknowledgements The authors would like to thank R. Mesiar for his valuable comments and inspiration. References [1] J.P. Bézivin, M.S. Tomás, On the determination of strict t-norms on some diagonal segments, Aequations Math. 45 (1993) 239–245. [2] J. Fodor, S. Ovchinnikov, personal communication, 2003. [3] G.H. Hardy, J.E. Littlewood, G. Pólya, Inequalities, 2nd Edition, Cambridge University Press, Cambridge, 1952. [4] C. Kimberling, On a class of associative functions, Publ. Math. Debrecen 20 (1973) 21–39. [5] E.P. Klement, R. Mesiar, E. Pap, Triangular Norms, Kluwer, Dordrecht, Boston, London, 2000. [6] E.P. Klement, R. Mesiar, E. Pap, Triangular norms: some open questions, Proc. Linz Seminar 2003, Johannes Kepler University Linz, 2003, pp. 135–138. [7] E.P. Klement, R. Mesiar, E. Pap, Problems on triangular norms and related operators, Fuzzy Sets and Systems 145 (2004) 471–479. [8] A. Marková, T-sum of L-R-fuzzy numbers, Fuzzy Sets and Systems 85 (1997) 379–384. [9] R. Mesiar, Shape preserving additions of fuzzy intervals, Fuzzy Sets and Systems 86 (1997) 73–78. [10] R. Mesiar, M. Navara, Diagonals of continuous triangular norms, Fuzzy Sets and Systems 104 (1999) 34–41. [11] R.B. Nelsen, An Introduction to Copulas, Springer, Berlin, 1999.