- Email: [email protected]
Local monotonicity properties of two-variable Gini means and the comparison theorem revisited
J. Math. Anal. Appl. 301 (2005) 427–438 www.elsevier.com/locate/jmaa
Local monotonicity properties of two-variable Gini means and the comparison theorem revisited ✩ Péter Czinder a , Zsolt Páles b,∗ a Berze Nagy János Grammar School, Kossuth str. 33, H-3200 Gyöngyös, Hungary b Institute of Mathematics, University of Debrecen, Pf. 12, H-4010 Debrecen, Hungary
Received 20 May 2004 Available online 30 September 2004 Submitted by B.S. Thomson
Abstract In this paper we examine local monotonicity properties of the so-called Gini means, defined by 1 x a + y a a−b if a = b, xb + yb Ga,b (x, y) = a a exp x log x + y log y if a = b xa + ya with respect to the parameters (a, b). As a consequence of our results, we obtain a new proof of the comparison theorem of two-variable Gini means. 2004 Elsevier Inc. All rights reserved. Keywords: Gini means; Local monotonicity properties; Comparison problem
✩
The research was supported by the Hungarian Research Fund (OTKA), Grant Nos. T-038072 and T-047373.
* Corresponding author.
E-mail addresses: [email protected] (P. Czinder), [email protected] (Z. Páles). 0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2004.08.006
428
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
1. Introduction. Preliminary results In a recent paper [2] we have studied an extension of the classical Hermite–Hadamard inequality (published first in [4]), which – in its original form – reads as follows: Theorem 1.1. Let J ⊂ R be an interval and f : J → R be a concave (convex) function. Then, for all subintervals [p, q] ⊂ J , q 1 p+q f (p) + f (q) . (1.1) f () q − p f (x) dx () 2 2 p
A generalization of this result was obtained in [2], where we proved that, for functions symmetric with respect to a point m ∈ J – that is, satisfying f (m − t) + f (m + t) = 2f (m) for all t ∈ (J − m) ∩ (m − J ) – the same inequality holds: Theorem 1.2. Let f : J → R be symmetric with respect to an element m ∈ J ; furthermore, suppose that f is convex over the interval J ∩ (−∞, m] and concave over J ∩ [m, ∞). Then, for any interval [p, q] ⊂ J , the inequality (1.1) is valid if p+q m. 2 () (In (1.1) the reversed inequalities hold if f is concave over the interval J ∩ (−∞, m] and convex over J ∩ [m, ∞).) This result was applied to the so-called two-variable Gini means to obtain lower and upper bound for the means of unequal parameters in terms of those with equal parameters. To restate this result here in details, we recall the definition of Gini means first. Given two real parameters a, b, if x, y are positive numbers, their Gini mean Ga,b (x, y) (cf. [3]) is defined by 1 x a + y a a−b if a = b, xb + yb Ga,b (x, y) = x a log x + y a log y exp if a = b. xa + ya Fixing the positive numbers x and y, the function gx,y : R → R+ ,
t → log Gt,t (x, y)
(1.2)
turns out to satisfy the assumptions of Theorem 1.2 (see [2] for the details), namely, gx,y is symmetric with respect to the point 0, convex over R− , and concave over R+ . Therefore, applying Theorem 1.2, the following result can be deduced: 0. Then Theorem 1.3. Let a, b be real numbers so that a + b () G (x, y) G (x, y)G (x, y) for any x, y > 0. (1.3) G a+b , a+b (x, y) () a,b a,a b,b () 2
2
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
429
In Section 2, by proving also monotonicity properties of the function gx,y , we obtain a new upper (lower) estimate for Ga,b (x, y) in terms of a weighted geometric mean of Ga,a (x, y) and Gb,b (x, y). Then, applying these inequalities, we will be able to determine the sign of the directional derivatives of the function Gx,y : R × R −→ R+ ,
(s, t) → Gs,t (x, y),
(1.4)
depending on the direction. Finally, in Section 3, we give a new form with a new proof of the comparison theorem of Gini means found by the second author in [6]. This new form covers also the case of equal parameters.
2. A variant of the Hermite–Hadamard inequality In the sequel, we shall need a new variant of (1.1), where the left hand side is replaced by a certain weighted arithmetic mean of f (p) and f (q). To state this result, we recall the notions of the positive and negative part functions defined, for x ∈ R, by x + := max(x, 0) =
|x| + x 2
and x − := max(−x, 0) =
|x| − x . 2
Lemma 2.1. Let f : J → R be symmetric with respect to an element m ∈ J , furthermore, suppose that f is increasing. Then, for any interval [p, q] ⊂ J ,
(q − m)+ − (p − m)+ f (q) + (p − m)− − (q − m)− f (p) ()
q f (x) dx
(2.1)
p
if
p+q m. 2 ()
Proof. We consider first the case (p + q)/2 m and assume that p m. Then m 2m − p q. Applying the symmetry of f with respect to the point m twice and using also its monotonicity, we have: q
m f (t) dt =
p
2m−p
f (t) dt + p
q
f (t) dt + m
2m−p
m−p
=
f (t) dt
q
f (m − t) + f (m + t) dt +
0
2m−p
q = 2(m − p)f (m) + 2m−p
f (t) dt
f (t) dt
430
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
= (m − p)f (p) + (m − p)f (2m − p) +
q f (t) dt
2m−p
(m − p)f (p) + (m − p)f (q) + (p + q − 2m)f (q) = (q − m)f (q) + (m − p)f (p)
= (q − m)+ − (p − m)+ f (q) + (p − m)− − (q − m)− f (p). If m p, then, using only the monotonicity of f , we get q f (t) dt (q − p)f (q) p
= (q − m)+ − (p − m)+ f (q) + (p − m)− − (q − m)− f (p).
In the case (p + q)/2 m, a similar argument completes the proof. 2 Remark 2.2. Using the monotonicity of f , it is elementary to see that the left-hand side of (2.1) can equivalently be written also in the form
p+q min (q − m)f (q) + (m − p)f (p), (q − p)f (q) if m. 2
(2.2)
Indeed, if p m, then
(q − m)+ − (p − m)+ f (q) + (p − m)− − (q − m)− f (p) = (q − m)f (q) + (m − p)f (p) and (q − m)f (q) + (m − p)f (p) (q − p)f (q) which results (2.2). In the case m p, we have that
(q − m)+ − (p − m)+ f (q) + (p − m)− − (q − m)− f (p) = (q − p)f (q) and (q − m)f (q) + (m − p)f (p) (q − p)f (q), which also leads to (2.2). Analogously, if (p + q)/2 m, then the left-hand side of (2.1) can be replaced by
max (q − m)f (q) + (m − p)f (p), (q − p)f (p) . Now we apply the result of Lemma 2.1 for Gini means to replace the left-hand side in (1.3) by a weighted geometric mean of Ga,a and Gb,b . Due to the symmetry of Ga,b (x, y) in its parameters – that is, the symmetry of the function Gx,y – it suffices to handle the situation b a in the sequel. 0 then, for all positive x, y, Theorem 2.3. If b < a and a + b ()
Ga,a (x, y)
a+ −b+
b− −a− a−b Ga,b (x, y). Gb,b (x, y) a−b ()
(2.3)
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
431
Proof. Using the result of F. Qi [7], we get that the function gx,y defined in (1.2) is strictly increasing for all fixed x, y > 0 with x = y. A cardinal property of the Gini means with different parameters is that their logarithm can be expressed as the integral average of the logarithm of Gini means with equal parameters, more precisely, the following formula holds (see, e.g., [2,5]): 1 log Ga,b (x, y) = a −b
a gx,y (t) dt
(x, y ∈ R+ ).
(2.4)
b
In the proof of (2.3), we can restrict ourselves to the upper direction in the inequalities; we could use an analogous argument for the reversed signs. Then, in view of Lemma 2.1, 1 log Ga,b (x, y) = a −b
a gx,y (t) dt b
a + − b+ b− − a − gx,y (a) + gx,y (b) a−b a−b
a + − b+ b− − a − log Ga,a (x, y) + log Gb,b (x, y), = a−b a−b which is equivalent to the desired inequality (2.3). 2 Combining the results of Theorems 1.3 and 2.3, we get the following lower and upper estimates for the Gini mean Ga,b in terms of a weighted geometric mean of Ga,a and Gb,b : Corollary 2.4. For all real a, b with b a, (a, b) = (0, 0) and for all positive x, y,
1
1 2 G 2 G b,b (x, y) a,b (x, y) Ga,a (x, y) if 0 b a;
1
1
a
−b (ii) Ga,a (x, y) 2 Gb,b (x, y) 2 Ga,b (x, y) Ga,a (x, y) a−b Gb,b (x, y) a−b (i)
(iii)
(iv)
Ga,a (x, y)
if 0 −b a;
1
1
a
−b Ga,a (x, y) 2 Gb,b (x, y) 2 Ga,b (x, y) Ga,a (x, y) a−b Gb,b (x, y) a−b
if 0 a −b;
1
1 Ga,a (x, y) 2 Gb,b (x, y) 2 Ga,b (x, y) Gb,b (x, y) if b a 0.
Proof. All the left-hand side inequalities follow from the right-hand side inequality (1.3) of Theorem 1.3. The right-hand side inequalities in (i)–(iv) are consequences of the inequality (2.3). 2 The importance of the inequalities established in Corollary 2.4 will be clear when we apply them to investigate the directional derivatives of the function Gx,y . Lemma 2.5. Suppose that b a and let the positive numbers x and y be fixed. The directional derivative of the function Gx,y in direction d = (d1 , d2 ) is nonnegative at the point (a, b) if and only if
432
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
Ga,a (x, y)
d1
−d
d −d Gb,b (x, y) 2 Ga,b (x, y) 1 2 ,
d1 + d2 0,
if a = b,
if a = b.
(2.5)
Proof. Then the directional derivative of Gx,y at (a, b) in direction d can be calculated as follows: ∂d Gx,y (a, b) = d1 · ∂1 Gx,y (a, b) + d2 · ∂2 Gx,y (a, b). (Here ∂i stands for the partial derivative with respect to the ith variable.) An easy calculation shows that, in the case a = b, Ga,a (x, y) Ga,b (x, y) log and a−b Ga,b (x, y) Ga,b (x, y) Ga,b (x, y) ∂2 Gx,y (a, b) = log . a−b Gb,b (x, y)
∂1 Gx,y (a, b) =
Therefore, ∂d Gx,y (a, b) =
Ga,b (x, y) (Ga,a (x, y))d1 (Gb,b (x, y))−d2 . log a −b (Ga,b (x, y))d1 −d2
By the assumption a > b, this expression is nonnegative if and only if
d
−d
d −d Ga,a (x, y) 1 Gb,b (x, y) 2 Ga,b (x, y) 1 2 , that is, the first inequality in (2.5) holds. In the case a = b, we have that log x − log y 2 (xy)a Ga,a (x, y). ∂1 Gx,y (a, a) = ∂2 Gx,y (a, a) = xa + ya Thus, it is easily seen that ∂d Gx,y (a, a) 0 if and only if d1 + d2 0.
2
Combining Lemma 2.5 and Corollary 2.4, we get the following result. Corollary 2.6. For b a with (a, b) = (0, 0), define the vectors ua,b and va,b in the following way: (1, 0), if 0 b a, (−1, 1), if 0 b a, b
1, a , (−1, 1), if 0 < −b a, if 0 < −b < a, ua,b = (2.6) va,b = a
if 0 < a < −b, (1, −1), if 0 < a −b, b,1 , (0, 1), if b a 0. (1, −1), if b a 0, Then the maps (a, b) → ua,b and (a, b) → va,b are continuous on the domain indicated, furthermore, the directional derivative of the function Gx,y , at the point (a, b) is nonnegative in the directions ua,b and va,b for all fixed positive numbers x, y. Proof. In view of Lemma 2.5, we have to check (2.5) for the vectors d = ua,b and d = va,b . By substituting these values of d into (2.5), it is easily seen that the conditions obtained are equivalent to the inequalities established in Corollary 2.4. 2 The following lemma offers a useful sufficient condition for the comparison of two Gini means.
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
433
Lemma 2.7. Let the positive numbers x and y be fixed and let (a, b), (c, d) be two arbitrary points in R2 . Suppose that the directional derivative of Gx,y is nonnegative in the direction (c − a, d − b) at any point of the segment
(a, b), (c, d) = a + t (c − a), b + t (d − b) t ∈ [0, 1] . Then Ga,b (x, y) Gc,d (x, y). Proof. Define
ϕx,y (t) := Ga+t (c−a),b+t (d−b)(x, y) = Gx,y a + t (c − a), b + t (d − b) (t ∈ [0, 1]).
Then ϕ is differentiable on R and
(t) = ∂1 Gx,y a + t (c − a), b + t (d − b) · (c − a) ϕx,y
+ ∂2 Gx,y a + t (c − a), b + t (d − b) · (d − b)
= ∂(c−a,d−b) Gx,y a + t (c − a), b + t (d − b) 0 by our assumption. Thus ϕx,y is nondecreasing on [0, 1], whence Ga,b (x, y) = ϕx,y (0) ϕx,y (1) = Gc,d (x, y).
2
3. Asymptotic properties of Gini means In order to obtain necessary conditions for the comparison of Gini means, we shall need the following lemma which establishes asymptotic properties of Gini means. Theorem 3.1. Let a, b ∈ R be arbitrary. Then Ga,b (t, 1) − t →1 (t − 1)2 lim
t +1 2
=
a+b−1 , 8
(3.1)
|a| − |b| log Ga,b (et , e−t ) lim = µ(a, b) := a−b t →∞ t sign(a) and t + log Ga,b (et , e−t ) 1 log = ν(a, b) := lim t →∞ 2t t − log Ga,b (et , e−t )
if a = b,
(3.2)
if a = b,
min(a, b)
if a, b 0, 0 if ab < 0, max(a, b) if a, b 0.
(3.3)
Proof. To prove (3.1), we apply L’Hospital’s rule twice: Ga,b (t, 1) − t →1 (t − 1)2 lim
t +1 2
∂1 Ga,b (t, 1) − t →1 2(t − 1) a+b−1 . = 8
= lim
1 2
∂12 Ga,b (t, 1) ∂12 Ga,b (1, 1) = t →1 2 2
= lim
434
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
If a = b, then 1
1
|a| − |b| log Ga,b (et , e−t ) log(eat + e−at ) t − log(ebt + e−bt ) t = lim = . t →∞ t →∞ t a−b a−b On the other hand, if a = b, then lim
log Ga,a (et , e−t ) eat log et + e−at log e−t = lim = lim tanh(at) = sign(a). t →∞ t →∞ t →∞ t t (eat + e−at ) lim
Thus (3.2) is also verified. For (3.3) it is enough to show that log(t + log Ga,b (et , e−t )) = lim t →∞ 2t and log(t − log Ga,b (et , e−t )) = lim t →∞ 2t
0 if a, b 0, 0 if ab < 0, max(a, b) if a, b 0,
(3.4)
− min(a, b) if a, b 0, 0 if ab < 0, 0 if a, b 0,
(3.5)
because (3.3) is just the difference of (3.4) and (3.5). In the proof of (3.4) we distinguish two cases. If either ab < 0 or a, b 0, then −1 < µ(a, b) 1. Thus, in view of (3.2), we get log(t + log Ga,b (et , e−t )) log t + log(1 + log Ga,b (et , e−t )/t) = lim t →∞ t →∞ 2t 2t log t log(1 + µ(a, b)) + lim = 0. = lim t →∞ 2t t →∞ 2t Finally, we consider the case a, b 0. We may assume that b a 0; we shall prove that lim
log(t + log Ga,b (et , e−t )) = a. t →∞ 2t If b < a, then a short calculation shows that lim
(3.6)
1 1 1 + e2at 1 t −t log t + log Ga,b (e , e ) = log log . 2t 2t a −b 1 + e2bt
To determine the limit of the last expression, we can apply L’Hospital’s rule. Computing the derivatives of the numerator and the denominator of the right hand side term, after some transformations, it turns out that their ratio is equal to a + (a − b)e2bt − be−2(a−b)t . 2at ) log(1+e2bt ) (1 + e2at )(1 + e2bt ) log(1+e − 2at 2at e e
(3.7)
In (3.7) the limit of the numerator equals a. With L’Hospital’s rule it can immediately be checked that 1 if p = q < 0, log(1 + e2pt ) log 2 if p = q = 0, (3.8) lim = t →∞ e2qt 0 if p < q 0,
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
435
which means that the limit of (3.7) in each cases equals a, as we stated. The case of the equal parameters in (3.4) can be treated similarly. The proof of (3.5) is completely analogous, therefore, it is omitted. 2 Remark 3.2. Using the homogeneity of the Gini means, one can obtain the following more general statements: Ga,b (x, y) − x+y a+b−1 2 (t > 0), = 2 (x,y)→(t,t ) (x − y) 8t
1 √
µ(a,b) lim Ga,b x t , y t t = xy max x/y, y/x
(3.9)
lim
t →∞
and
lim
t →∞
log y t − log Ga,b (x t , y t ) log Ga,b (x t , y t ) − log x t
(x, y > 0),
(3.10)
1 t
= (x/y)ν(a,b)
(x, y > 0).
(3.11)
Observe that with x = e, y = 1/e and after taking the logarithm of both sides, the relations (3.10) and (3.11) reduce to (3.2) and (3.3), respectively. Conversely, using the homogeneity of Gini means, (3.9), (3.10), and (3.11) can easily be deduced from (3.1), (3.2), and (3.3).
4. Comparison of Gini means In the result below, we restate the necessary and sufficient conditions for the comparison of two-variable Gini means obtained in [6] (and also in [1]) in a completely new form with a completely new proof. Theorem 4.1. Let a, b, c, d ∈ R be arbitrary parameters. Then Ga,b (x, y) Gc,d (x, y)
(4.1)
holds for all positive x and y if and only if a, b, c, d satisfy the following three conditions: a + b c + d,
(4.2)
µ(a, b) µ(c, d),
(4.3)
ν(a, b) ν(c, d).
(4.4)
Proof. Necessity. Assume that (4.1) holds for all x, y > 0. Then, substituting x = t, y = 1, we get that Ga,b (t, 1) − (t − 1)2
t +1 2
Gc,d (t, 1) − (t − 1)2
t +1 2
(t > 0, t = 1).
Now, taking the limit t → 1 and using (3.1) of Theorem 3.1, we obtain a +b−1 c+d −1 , 8 8 which is equivalent to (4.2).
436
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
The substitution x = et , y = e−t results from (4.1) that log Ga,b (et , e−t ) log Gc,d (et , e−t ) (t > 0). t t Therefore, taking the limit t → ∞ and using Theorem 3.1, we get the necessity of (4.3). Finally, again with the substitution x = et , y = e−t we have (by the strict mean value property of the Gini means) that
−t < log Ga,b et , e−t log Gc,d et , e−t < t (t > 0), therefore, 0<
t + log Ga,b (et , e−t ) t + log Gc,d (et , e−t ) t − log Ga,b (et , e−t ) t − log Gc,d (et , e−t )
(t > 0).
Hence, in view of Theorem 3.1, t + log Ga,b (et , e−t ) t + log Gc,d (et , e−t ) 1 1 log lim log t −t t →∞ 2t t − log Ga,b (e , e ) t →∞ 2t t − log Gc,d (et , e−t ) = ν(c, d)
ν(a, b) = lim
and thus the necessity of (4.4) is also proved. Sufficiency. Assume that conditions (4.2)–(4.4) hold and let x and y be two arbitrary positive numbers throughout the proof. In order to distinguish various cases according to the positions of the points (a, b) and (c, d), consider the following five subsets of the half-plain H := {(s, t) | t s}:
H1 := (s, t) | t s 0, (s, t) = (0, 0) , H2 := (s, t) | 0 < s < −t ,
H3 := (s, t) ∈ H | s + t = 0 ,
H4 := (s, t) | 0 −t < s , H5 := (s, t) | 0 < t s . Evidently H = H1 ∪ H2 ∪ H3 ∪ H4 ∪ H5 , furthermore, H1 ∪ H2 = {(s, t) ∈ H | s + t < 0}, H4 ∪ H5 = {(s, t) ∈ H | s + t > 0}, and Hi ∩ Hj = ∅ for all 1 i < j 5. By the symmetry of the parameters, we can assume that a b and c d, that is, (a, b), (c, d) ∈ H . Due to the pairwise disjointness of the sets H1 , H2 , H3 , H4 , and H5 , there exist unique indices i, j such that (a, b) ∈ Hi and (c, d) ∈ Hj . Applying conditions (4.2)–(4.4), we prove first that i j holds. If (a, b) ∈ H1 , then there is nothing to prove. If (a, b) ∈ H2 , then a > 0, hence µ(a, b) > −1. Thus, condition (4.3) yields that µ(c, d) > −1, hence, (c, d) cannot be in H1 , i.e., it belongs to H2 ∪ H3 ∪ H4 ∪ H5 . In the case when (a, b) ∈ H3 , we have that a + b = 0. Hence, by (4.2), it follows that c + d 0, i.e., (c, d) ∈ H3 ∪ H4 ∪ H5 . If (a, b) ∈ H4 , then a + b > 0. Thus (4.2) yields that c + d > 0, which is equivalent to (c, d) ∈ H4 ∪ H5 . Finally, if (a, b) is in H5 , then 0 < b. Therefore ν(a, b) = min(a, b) > 0, whence, in view of condition (4.4), we get that ν(c, d) > 0. Thus (c, d) has to be an element of H5 , too. In the rest of the proof, we may assume that (a, b) ∈ Hi and (c, d) ∈ Hj for some indices 1 i j 5. Our aim is to prove that Ga,b (x, y) Gp,q (x, y) Gc,d (x, y),
(4.5)
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
437
where (p, q) is defined by distinguishing five cases (i)–(v) according to the possible positions of the points (a, b) and (c, d) in H as follows: (i) (0, 0), if (a, b) ∈ H1 ∪ H2 ∪ H3 , (c, d) ∈ H3 ∪ H4 ∪ H5 ; (ii) (c, a + b − c), if (a, b) ∈ H1 , (c, d) ∈ H1 ; a+b a+b c, d , if (a, b) ∈ H1 ∪ H2 , (c, d) ∈ H2 ; (p, q) := (iii) c+d c+d c+d c+d (iv) a, b , if (a, b) ∈ H4 , (c, d) ∈ H4 ∪ H5 ; a+b a+b (v) (c + d − b, b), if (a, b) ∈ H5 , (c, d) ∈ H5 . For, we will show that at any point of the two segments [(a, b), (p, q)] and [(p, q), (c, d)] the directional derivatives of Gx,y , defined by the vectors (p − a, q − b) and (c − p, d − q), respectively, are nonnegative. Thus, in view of Lemma 2.7, the desired inequality (4.5) results, which demonstrates the statement. Case (i). We have to show that Ga,b (x, y) G0,0 (x, y) Gc,d (x, y) if a + b 0 c + d. Due to the symmetry, we may deal only with the first inequality. If (a, b) ∈ H1 , then the direction (p − a, q − b) = (−a, −b) = (−a − b)(0, 1) + (−a)(1, −1) is a cone combination of the vectors u = (1, −1) and v = (0, 1). Since the segment [(a, b), (0, 0)] is contained in H1 ∪ {(0, 0)}, hence, by Corollary 2.6, the directional derivative of Gx,y in the direction (−a, −b) is nonnegative at any point of [(a, b), (0, 0)]. Thus, Lemma 2.7 results Ga,b (x, y) G0,0 (x, y). Similarly, if (a, b) ∈ H2 , then (p − a, q − b) = (−a, −b) = (−b)(a/b, 1). Therefore, by Corollary 2.6, the directional derivative of Gx,y in the direction (−a, −b) is nonnegative at any point of the segment [(a, b), (0, 0)] ⊂ H2 ∪ {(0, 0)}. Applying Lemma 2.7, Ga,b (x, y) G0,0 (x, y) follows again. Finally, if (a, b) ∈ H3 , then an elementary computation shows that the identity Ga,b (x, y) = G0,0 (x, y) holds. Case (ii). Since (c, d) ∈ H1 , we know that c 0, and due to condition (4.2), we get that a + b − c d 0. Thus, (p, q) ∈ H1 , whence, by the convexity of H1 , it follows that [(a, b), (p, q)] ⊂ H1 . By condition (4.4), we have that a c, hence (p − a, q − b) = (c − a, a − c) is obtained from u = (1, −1) by multiplication with a nonnegative scalar. Therefore, by Corollary 2.6, the directional derivative of Gx,y in the direction u is nonnegative at any point of the segment [(a, b), (p, q)]. Applying Lemma 2.7, we obtain that Ga,b (x, y) Gp,q (x, y). We have a similar situation for the segment [(p, q), (c, d)] ⊂ H1 . In this case, the vector (c − p, d − q) = (0, c + d − a − b) is co-directional with v = (0, 1) (by condition (4.2)). Using Corollary 2.6 again, the directional derivative of Gx,y in direction v is nonnegative at any point of the segment [(p, q), (c, d)]. Therefore, by Lemma 2.7, we get that Gp,q (x, y) Gc,d (x, y). Case (iii). Then (c, d) ∈ H2 yields that c > 0 and d < 0. The inequality ad − bc 0 is obviously valid if (a, b) ∈ H1 , because then a and b are nonpositive. In the case (a, b) ∈ H2 ,
438
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
we have that b < 0 < a, therefore, the condition (4.3) ensures that ad − bc 0 also holds. Thus the direction bc − ad (1, −1) (p − a, q − b) = c+d is co-directional with the vector u = (1, −1). Using Corollary 2.6, we can see that the directional derivative of Gx,y , in the direction u is nonnegative at any point of the segment [(a, b), (p, q)] ⊂ H1 ∪ H2 . Hence, by Lemma 2.7, it follows that Ga,b (x, y) Gp,q (x, y). Observe that (p, q) ∈ H2 , hence we have that the segment [(p, q), (c, d)] is also contained in H2 . On the other hand, (c + d − a − b)d c ,1 , (c − p, d − q) = c+d d
hence (c − p, d − q) is co-directional with the vector v = dc , 1 . In view of Corollary 2.6 again, we obtain that the directional derivative of Gx,y in direction v is nonnegative at any point of the segment [(p, q), (c, d)] ⊂ H2 . Therefore, by Lemma 2.7, we get that Gp,q (x, y) Gc,d (x, y). The Cases (iv), (v) are completely analogous to Cases (iii), (ii), respectively. Therefore, the details are left to the reader. 2
References [1] P. Czinder, Zs. Páles, A general Minkowski-type inequality for two variable Gini means, Publ. Math. Debrecen 57 (2000) 203–216. [2] P. Czinder, Zs. Páles, An extension of the Hermite–Hadamard inequality and an application for Gini and Stolarsky means, J. Inequal. Pure Appl. Math. 5 (2004), Article 42, 8 p. (electronic). [3] C. Gini, Di una formula compressiva delle medie, Metron 13 (1938) 3–22. [4] J. Hadamard, Étude sur les propriétés des fonctions entières et en particulier d’une fonction considérée par Riemann, J. Math. Pures Appl. 58 (1893) 171–215. [5] E. Neumann, J. Sándor, Inequalities involving Stolarsky and Gini means, Math. Pannon. 14 (2003) 29–44. [6] Zs. Páles, Inequalities for sums of powers, J. Math. Anal. Appl. 131 (1988) 265–270. [7] F. Qi, Generalized abstracted mean values, J. Inequal. Pure Appl. Math. 1 (2000), Article 4 (electronic).
Local monotonicity properties of two-variable Gini means and the comparison theorem revisited ✩ Péter Czinder a , Zsolt Páles b,∗ a Berze Nagy János Grammar School, Kossuth str. 33, H-3200 Gyöngyös, Hungary b Institute of Mathematics, University of Debrecen, Pf. 12, H-4010 Debrecen, Hungary
Received 20 May 2004 Available online 30 September 2004 Submitted by B.S. Thomson
Abstract In this paper we examine local monotonicity properties of the so-called Gini means, defined by 1 x a + y a a−b if a = b, xb + yb Ga,b (x, y) = a a exp x log x + y log y if a = b xa + ya with respect to the parameters (a, b). As a consequence of our results, we obtain a new proof of the comparison theorem of two-variable Gini means. 2004 Elsevier Inc. All rights reserved. Keywords: Gini means; Local monotonicity properties; Comparison problem
✩
The research was supported by the Hungarian Research Fund (OTKA), Grant Nos. T-038072 and T-047373.
* Corresponding author.
E-mail addresses: [email protected] (P. Czinder), [email protected] (Z. Páles). 0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2004.08.006
428
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
1. Introduction. Preliminary results In a recent paper [2] we have studied an extension of the classical Hermite–Hadamard inequality (published first in [4]), which – in its original form – reads as follows: Theorem 1.1. Let J ⊂ R be an interval and f : J → R be a concave (convex) function. Then, for all subintervals [p, q] ⊂ J , q 1 p+q f (p) + f (q) . (1.1) f () q − p f (x) dx () 2 2 p
A generalization of this result was obtained in [2], where we proved that, for functions symmetric with respect to a point m ∈ J – that is, satisfying f (m − t) + f (m + t) = 2f (m) for all t ∈ (J − m) ∩ (m − J ) – the same inequality holds: Theorem 1.2. Let f : J → R be symmetric with respect to an element m ∈ J ; furthermore, suppose that f is convex over the interval J ∩ (−∞, m] and concave over J ∩ [m, ∞). Then, for any interval [p, q] ⊂ J , the inequality (1.1) is valid if p+q m. 2 () (In (1.1) the reversed inequalities hold if f is concave over the interval J ∩ (−∞, m] and convex over J ∩ [m, ∞).) This result was applied to the so-called two-variable Gini means to obtain lower and upper bound for the means of unequal parameters in terms of those with equal parameters. To restate this result here in details, we recall the definition of Gini means first. Given two real parameters a, b, if x, y are positive numbers, their Gini mean Ga,b (x, y) (cf. [3]) is defined by 1 x a + y a a−b if a = b, xb + yb Ga,b (x, y) = x a log x + y a log y exp if a = b. xa + ya Fixing the positive numbers x and y, the function gx,y : R → R+ ,
t → log Gt,t (x, y)
(1.2)
turns out to satisfy the assumptions of Theorem 1.2 (see [2] for the details), namely, gx,y is symmetric with respect to the point 0, convex over R− , and concave over R+ . Therefore, applying Theorem 1.2, the following result can be deduced: 0. Then Theorem 1.3. Let a, b be real numbers so that a + b () G (x, y) G (x, y)G (x, y) for any x, y > 0. (1.3) G a+b , a+b (x, y) () a,b a,a b,b () 2
2
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
429
In Section 2, by proving also monotonicity properties of the function gx,y , we obtain a new upper (lower) estimate for Ga,b (x, y) in terms of a weighted geometric mean of Ga,a (x, y) and Gb,b (x, y). Then, applying these inequalities, we will be able to determine the sign of the directional derivatives of the function Gx,y : R × R −→ R+ ,
(s, t) → Gs,t (x, y),
(1.4)
depending on the direction. Finally, in Section 3, we give a new form with a new proof of the comparison theorem of Gini means found by the second author in [6]. This new form covers also the case of equal parameters.
2. A variant of the Hermite–Hadamard inequality In the sequel, we shall need a new variant of (1.1), where the left hand side is replaced by a certain weighted arithmetic mean of f (p) and f (q). To state this result, we recall the notions of the positive and negative part functions defined, for x ∈ R, by x + := max(x, 0) =
|x| + x 2
and x − := max(−x, 0) =
|x| − x . 2
Lemma 2.1. Let f : J → R be symmetric with respect to an element m ∈ J , furthermore, suppose that f is increasing. Then, for any interval [p, q] ⊂ J ,
(q − m)+ − (p − m)+ f (q) + (p − m)− − (q − m)− f (p) ()
q f (x) dx
(2.1)
p
if
p+q m. 2 ()
Proof. We consider first the case (p + q)/2 m and assume that p m. Then m 2m − p q. Applying the symmetry of f with respect to the point m twice and using also its monotonicity, we have: q
m f (t) dt =
p
2m−p
f (t) dt + p
q
f (t) dt + m
2m−p
m−p
=
f (t) dt
q
f (m − t) + f (m + t) dt +
0
2m−p
q = 2(m − p)f (m) + 2m−p
f (t) dt
f (t) dt
430
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
= (m − p)f (p) + (m − p)f (2m − p) +
q f (t) dt
2m−p
(m − p)f (p) + (m − p)f (q) + (p + q − 2m)f (q) = (q − m)f (q) + (m − p)f (p)
= (q − m)+ − (p − m)+ f (q) + (p − m)− − (q − m)− f (p). If m p, then, using only the monotonicity of f , we get q f (t) dt (q − p)f (q) p
= (q − m)+ − (p − m)+ f (q) + (p − m)− − (q − m)− f (p).
In the case (p + q)/2 m, a similar argument completes the proof. 2 Remark 2.2. Using the monotonicity of f , it is elementary to see that the left-hand side of (2.1) can equivalently be written also in the form
p+q min (q − m)f (q) + (m − p)f (p), (q − p)f (q) if m. 2
(2.2)
Indeed, if p m, then
(q − m)+ − (p − m)+ f (q) + (p − m)− − (q − m)− f (p) = (q − m)f (q) + (m − p)f (p) and (q − m)f (q) + (m − p)f (p) (q − p)f (q) which results (2.2). In the case m p, we have that
(q − m)+ − (p − m)+ f (q) + (p − m)− − (q − m)− f (p) = (q − p)f (q) and (q − m)f (q) + (m − p)f (p) (q − p)f (q), which also leads to (2.2). Analogously, if (p + q)/2 m, then the left-hand side of (2.1) can be replaced by
max (q − m)f (q) + (m − p)f (p), (q − p)f (p) . Now we apply the result of Lemma 2.1 for Gini means to replace the left-hand side in (1.3) by a weighted geometric mean of Ga,a and Gb,b . Due to the symmetry of Ga,b (x, y) in its parameters – that is, the symmetry of the function Gx,y – it suffices to handle the situation b a in the sequel. 0 then, for all positive x, y, Theorem 2.3. If b < a and a + b ()
Ga,a (x, y)
a+ −b+
b− −a− a−b Ga,b (x, y). Gb,b (x, y) a−b ()
(2.3)
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
431
Proof. Using the result of F. Qi [7], we get that the function gx,y defined in (1.2) is strictly increasing for all fixed x, y > 0 with x = y. A cardinal property of the Gini means with different parameters is that their logarithm can be expressed as the integral average of the logarithm of Gini means with equal parameters, more precisely, the following formula holds (see, e.g., [2,5]): 1 log Ga,b (x, y) = a −b
a gx,y (t) dt
(x, y ∈ R+ ).
(2.4)
b
In the proof of (2.3), we can restrict ourselves to the upper direction in the inequalities; we could use an analogous argument for the reversed signs. Then, in view of Lemma 2.1, 1 log Ga,b (x, y) = a −b
a gx,y (t) dt b
a + − b+ b− − a − gx,y (a) + gx,y (b) a−b a−b
a + − b+ b− − a − log Ga,a (x, y) + log Gb,b (x, y), = a−b a−b which is equivalent to the desired inequality (2.3). 2 Combining the results of Theorems 1.3 and 2.3, we get the following lower and upper estimates for the Gini mean Ga,b in terms of a weighted geometric mean of Ga,a and Gb,b : Corollary 2.4. For all real a, b with b a, (a, b) = (0, 0) and for all positive x, y,
1
1 2 G 2 G b,b (x, y) a,b (x, y) Ga,a (x, y) if 0 b a;
1
1
a
−b (ii) Ga,a (x, y) 2 Gb,b (x, y) 2 Ga,b (x, y) Ga,a (x, y) a−b Gb,b (x, y) a−b (i)
(iii)
(iv)
Ga,a (x, y)
if 0 −b a;
1
1
a
−b Ga,a (x, y) 2 Gb,b (x, y) 2 Ga,b (x, y) Ga,a (x, y) a−b Gb,b (x, y) a−b
if 0 a −b;
1
1 Ga,a (x, y) 2 Gb,b (x, y) 2 Ga,b (x, y) Gb,b (x, y) if b a 0.
Proof. All the left-hand side inequalities follow from the right-hand side inequality (1.3) of Theorem 1.3. The right-hand side inequalities in (i)–(iv) are consequences of the inequality (2.3). 2 The importance of the inequalities established in Corollary 2.4 will be clear when we apply them to investigate the directional derivatives of the function Gx,y . Lemma 2.5. Suppose that b a and let the positive numbers x and y be fixed. The directional derivative of the function Gx,y in direction d = (d1 , d2 ) is nonnegative at the point (a, b) if and only if
432
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
Ga,a (x, y)
d1
−d
d −d Gb,b (x, y) 2 Ga,b (x, y) 1 2 ,
d1 + d2 0,
if a = b,
if a = b.
(2.5)
Proof. Then the directional derivative of Gx,y at (a, b) in direction d can be calculated as follows: ∂d Gx,y (a, b) = d1 · ∂1 Gx,y (a, b) + d2 · ∂2 Gx,y (a, b). (Here ∂i stands for the partial derivative with respect to the ith variable.) An easy calculation shows that, in the case a = b, Ga,a (x, y) Ga,b (x, y) log and a−b Ga,b (x, y) Ga,b (x, y) Ga,b (x, y) ∂2 Gx,y (a, b) = log . a−b Gb,b (x, y)
∂1 Gx,y (a, b) =
Therefore, ∂d Gx,y (a, b) =
Ga,b (x, y) (Ga,a (x, y))d1 (Gb,b (x, y))−d2 . log a −b (Ga,b (x, y))d1 −d2
By the assumption a > b, this expression is nonnegative if and only if
d
−d
d −d Ga,a (x, y) 1 Gb,b (x, y) 2 Ga,b (x, y) 1 2 , that is, the first inequality in (2.5) holds. In the case a = b, we have that log x − log y 2 (xy)a Ga,a (x, y). ∂1 Gx,y (a, a) = ∂2 Gx,y (a, a) = xa + ya Thus, it is easily seen that ∂d Gx,y (a, a) 0 if and only if d1 + d2 0.
2
Combining Lemma 2.5 and Corollary 2.4, we get the following result. Corollary 2.6. For b a with (a, b) = (0, 0), define the vectors ua,b and va,b in the following way: (1, 0), if 0 b a, (−1, 1), if 0 b a, b
1, a , (−1, 1), if 0 < −b a, if 0 < −b < a, ua,b = (2.6) va,b = a
if 0 < a < −b, (1, −1), if 0 < a −b, b,1 , (0, 1), if b a 0. (1, −1), if b a 0, Then the maps (a, b) → ua,b and (a, b) → va,b are continuous on the domain indicated, furthermore, the directional derivative of the function Gx,y , at the point (a, b) is nonnegative in the directions ua,b and va,b for all fixed positive numbers x, y. Proof. In view of Lemma 2.5, we have to check (2.5) for the vectors d = ua,b and d = va,b . By substituting these values of d into (2.5), it is easily seen that the conditions obtained are equivalent to the inequalities established in Corollary 2.4. 2 The following lemma offers a useful sufficient condition for the comparison of two Gini means.
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
433
Lemma 2.7. Let the positive numbers x and y be fixed and let (a, b), (c, d) be two arbitrary points in R2 . Suppose that the directional derivative of Gx,y is nonnegative in the direction (c − a, d − b) at any point of the segment
(a, b), (c, d) = a + t (c − a), b + t (d − b) t ∈ [0, 1] . Then Ga,b (x, y) Gc,d (x, y). Proof. Define
ϕx,y (t) := Ga+t (c−a),b+t (d−b)(x, y) = Gx,y a + t (c − a), b + t (d − b) (t ∈ [0, 1]).
Then ϕ is differentiable on R and
(t) = ∂1 Gx,y a + t (c − a), b + t (d − b) · (c − a) ϕx,y
+ ∂2 Gx,y a + t (c − a), b + t (d − b) · (d − b)
= ∂(c−a,d−b) Gx,y a + t (c − a), b + t (d − b) 0 by our assumption. Thus ϕx,y is nondecreasing on [0, 1], whence Ga,b (x, y) = ϕx,y (0) ϕx,y (1) = Gc,d (x, y).
2
3. Asymptotic properties of Gini means In order to obtain necessary conditions for the comparison of Gini means, we shall need the following lemma which establishes asymptotic properties of Gini means. Theorem 3.1. Let a, b ∈ R be arbitrary. Then Ga,b (t, 1) − t →1 (t − 1)2 lim
t +1 2
=
a+b−1 , 8
(3.1)
|a| − |b| log Ga,b (et , e−t ) lim = µ(a, b) := a−b t →∞ t sign(a) and t + log Ga,b (et , e−t ) 1 log = ν(a, b) := lim t →∞ 2t t − log Ga,b (et , e−t )
if a = b,
(3.2)
if a = b,
min(a, b)
if a, b 0, 0 if ab < 0, max(a, b) if a, b 0.
(3.3)
Proof. To prove (3.1), we apply L’Hospital’s rule twice: Ga,b (t, 1) − t →1 (t − 1)2 lim
t +1 2
∂1 Ga,b (t, 1) − t →1 2(t − 1) a+b−1 . = 8
= lim
1 2
∂12 Ga,b (t, 1) ∂12 Ga,b (1, 1) = t →1 2 2
= lim
434
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
If a = b, then 1
1
|a| − |b| log Ga,b (et , e−t ) log(eat + e−at ) t − log(ebt + e−bt ) t = lim = . t →∞ t →∞ t a−b a−b On the other hand, if a = b, then lim
log Ga,a (et , e−t ) eat log et + e−at log e−t = lim = lim tanh(at) = sign(a). t →∞ t →∞ t →∞ t t (eat + e−at ) lim
Thus (3.2) is also verified. For (3.3) it is enough to show that log(t + log Ga,b (et , e−t )) = lim t →∞ 2t and log(t − log Ga,b (et , e−t )) = lim t →∞ 2t
0 if a, b 0, 0 if ab < 0, max(a, b) if a, b 0,
(3.4)
− min(a, b) if a, b 0, 0 if ab < 0, 0 if a, b 0,
(3.5)
because (3.3) is just the difference of (3.4) and (3.5). In the proof of (3.4) we distinguish two cases. If either ab < 0 or a, b 0, then −1 < µ(a, b) 1. Thus, in view of (3.2), we get log(t + log Ga,b (et , e−t )) log t + log(1 + log Ga,b (et , e−t )/t) = lim t →∞ t →∞ 2t 2t log t log(1 + µ(a, b)) + lim = 0. = lim t →∞ 2t t →∞ 2t Finally, we consider the case a, b 0. We may assume that b a 0; we shall prove that lim
log(t + log Ga,b (et , e−t )) = a. t →∞ 2t If b < a, then a short calculation shows that lim
(3.6)
1 1 1 + e2at 1 t −t log t + log Ga,b (e , e ) = log log . 2t 2t a −b 1 + e2bt
To determine the limit of the last expression, we can apply L’Hospital’s rule. Computing the derivatives of the numerator and the denominator of the right hand side term, after some transformations, it turns out that their ratio is equal to a + (a − b)e2bt − be−2(a−b)t . 2at ) log(1+e2bt ) (1 + e2at )(1 + e2bt ) log(1+e − 2at 2at e e
(3.7)
In (3.7) the limit of the numerator equals a. With L’Hospital’s rule it can immediately be checked that 1 if p = q < 0, log(1 + e2pt ) log 2 if p = q = 0, (3.8) lim = t →∞ e2qt 0 if p < q 0,
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
435
which means that the limit of (3.7) in each cases equals a, as we stated. The case of the equal parameters in (3.4) can be treated similarly. The proof of (3.5) is completely analogous, therefore, it is omitted. 2 Remark 3.2. Using the homogeneity of the Gini means, one can obtain the following more general statements: Ga,b (x, y) − x+y a+b−1 2 (t > 0), = 2 (x,y)→(t,t ) (x − y) 8t
1 √
µ(a,b) lim Ga,b x t , y t t = xy max x/y, y/x
(3.9)
lim
t →∞
and
lim
t →∞
log y t − log Ga,b (x t , y t ) log Ga,b (x t , y t ) − log x t
(x, y > 0),
(3.10)
1 t
= (x/y)ν(a,b)
(x, y > 0).
(3.11)
Observe that with x = e, y = 1/e and after taking the logarithm of both sides, the relations (3.10) and (3.11) reduce to (3.2) and (3.3), respectively. Conversely, using the homogeneity of Gini means, (3.9), (3.10), and (3.11) can easily be deduced from (3.1), (3.2), and (3.3).
4. Comparison of Gini means In the result below, we restate the necessary and sufficient conditions for the comparison of two-variable Gini means obtained in [6] (and also in [1]) in a completely new form with a completely new proof. Theorem 4.1. Let a, b, c, d ∈ R be arbitrary parameters. Then Ga,b (x, y) Gc,d (x, y)
(4.1)
holds for all positive x and y if and only if a, b, c, d satisfy the following three conditions: a + b c + d,
(4.2)
µ(a, b) µ(c, d),
(4.3)
ν(a, b) ν(c, d).
(4.4)
Proof. Necessity. Assume that (4.1) holds for all x, y > 0. Then, substituting x = t, y = 1, we get that Ga,b (t, 1) − (t − 1)2
t +1 2
Gc,d (t, 1) − (t − 1)2
t +1 2
(t > 0, t = 1).
Now, taking the limit t → 1 and using (3.1) of Theorem 3.1, we obtain a +b−1 c+d −1 , 8 8 which is equivalent to (4.2).
436
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
The substitution x = et , y = e−t results from (4.1) that log Ga,b (et , e−t ) log Gc,d (et , e−t ) (t > 0). t t Therefore, taking the limit t → ∞ and using Theorem 3.1, we get the necessity of (4.3). Finally, again with the substitution x = et , y = e−t we have (by the strict mean value property of the Gini means) that
−t < log Ga,b et , e−t log Gc,d et , e−t < t (t > 0), therefore, 0<
t + log Ga,b (et , e−t ) t + log Gc,d (et , e−t ) t − log Ga,b (et , e−t ) t − log Gc,d (et , e−t )
(t > 0).
Hence, in view of Theorem 3.1, t + log Ga,b (et , e−t ) t + log Gc,d (et , e−t ) 1 1 log lim log t −t t →∞ 2t t − log Ga,b (e , e ) t →∞ 2t t − log Gc,d (et , e−t ) = ν(c, d)
ν(a, b) = lim
and thus the necessity of (4.4) is also proved. Sufficiency. Assume that conditions (4.2)–(4.4) hold and let x and y be two arbitrary positive numbers throughout the proof. In order to distinguish various cases according to the positions of the points (a, b) and (c, d), consider the following five subsets of the half-plain H := {(s, t) | t s}:
H1 := (s, t) | t s 0, (s, t) = (0, 0) , H2 := (s, t) | 0 < s < −t ,
H3 := (s, t) ∈ H | s + t = 0 ,
H4 := (s, t) | 0 −t < s , H5 := (s, t) | 0 < t s . Evidently H = H1 ∪ H2 ∪ H3 ∪ H4 ∪ H5 , furthermore, H1 ∪ H2 = {(s, t) ∈ H | s + t < 0}, H4 ∪ H5 = {(s, t) ∈ H | s + t > 0}, and Hi ∩ Hj = ∅ for all 1 i < j 5. By the symmetry of the parameters, we can assume that a b and c d, that is, (a, b), (c, d) ∈ H . Due to the pairwise disjointness of the sets H1 , H2 , H3 , H4 , and H5 , there exist unique indices i, j such that (a, b) ∈ Hi and (c, d) ∈ Hj . Applying conditions (4.2)–(4.4), we prove first that i j holds. If (a, b) ∈ H1 , then there is nothing to prove. If (a, b) ∈ H2 , then a > 0, hence µ(a, b) > −1. Thus, condition (4.3) yields that µ(c, d) > −1, hence, (c, d) cannot be in H1 , i.e., it belongs to H2 ∪ H3 ∪ H4 ∪ H5 . In the case when (a, b) ∈ H3 , we have that a + b = 0. Hence, by (4.2), it follows that c + d 0, i.e., (c, d) ∈ H3 ∪ H4 ∪ H5 . If (a, b) ∈ H4 , then a + b > 0. Thus (4.2) yields that c + d > 0, which is equivalent to (c, d) ∈ H4 ∪ H5 . Finally, if (a, b) is in H5 , then 0 < b. Therefore ν(a, b) = min(a, b) > 0, whence, in view of condition (4.4), we get that ν(c, d) > 0. Thus (c, d) has to be an element of H5 , too. In the rest of the proof, we may assume that (a, b) ∈ Hi and (c, d) ∈ Hj for some indices 1 i j 5. Our aim is to prove that Ga,b (x, y) Gp,q (x, y) Gc,d (x, y),
(4.5)
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
437
where (p, q) is defined by distinguishing five cases (i)–(v) according to the possible positions of the points (a, b) and (c, d) in H as follows: (i) (0, 0), if (a, b) ∈ H1 ∪ H2 ∪ H3 , (c, d) ∈ H3 ∪ H4 ∪ H5 ; (ii) (c, a + b − c), if (a, b) ∈ H1 , (c, d) ∈ H1 ; a+b a+b c, d , if (a, b) ∈ H1 ∪ H2 , (c, d) ∈ H2 ; (p, q) := (iii) c+d c+d c+d c+d (iv) a, b , if (a, b) ∈ H4 , (c, d) ∈ H4 ∪ H5 ; a+b a+b (v) (c + d − b, b), if (a, b) ∈ H5 , (c, d) ∈ H5 . For, we will show that at any point of the two segments [(a, b), (p, q)] and [(p, q), (c, d)] the directional derivatives of Gx,y , defined by the vectors (p − a, q − b) and (c − p, d − q), respectively, are nonnegative. Thus, in view of Lemma 2.7, the desired inequality (4.5) results, which demonstrates the statement. Case (i). We have to show that Ga,b (x, y) G0,0 (x, y) Gc,d (x, y) if a + b 0 c + d. Due to the symmetry, we may deal only with the first inequality. If (a, b) ∈ H1 , then the direction (p − a, q − b) = (−a, −b) = (−a − b)(0, 1) + (−a)(1, −1) is a cone combination of the vectors u = (1, −1) and v = (0, 1). Since the segment [(a, b), (0, 0)] is contained in H1 ∪ {(0, 0)}, hence, by Corollary 2.6, the directional derivative of Gx,y in the direction (−a, −b) is nonnegative at any point of [(a, b), (0, 0)]. Thus, Lemma 2.7 results Ga,b (x, y) G0,0 (x, y). Similarly, if (a, b) ∈ H2 , then (p − a, q − b) = (−a, −b) = (−b)(a/b, 1). Therefore, by Corollary 2.6, the directional derivative of Gx,y in the direction (−a, −b) is nonnegative at any point of the segment [(a, b), (0, 0)] ⊂ H2 ∪ {(0, 0)}. Applying Lemma 2.7, Ga,b (x, y) G0,0 (x, y) follows again. Finally, if (a, b) ∈ H3 , then an elementary computation shows that the identity Ga,b (x, y) = G0,0 (x, y) holds. Case (ii). Since (c, d) ∈ H1 , we know that c 0, and due to condition (4.2), we get that a + b − c d 0. Thus, (p, q) ∈ H1 , whence, by the convexity of H1 , it follows that [(a, b), (p, q)] ⊂ H1 . By condition (4.4), we have that a c, hence (p − a, q − b) = (c − a, a − c) is obtained from u = (1, −1) by multiplication with a nonnegative scalar. Therefore, by Corollary 2.6, the directional derivative of Gx,y in the direction u is nonnegative at any point of the segment [(a, b), (p, q)]. Applying Lemma 2.7, we obtain that Ga,b (x, y) Gp,q (x, y). We have a similar situation for the segment [(p, q), (c, d)] ⊂ H1 . In this case, the vector (c − p, d − q) = (0, c + d − a − b) is co-directional with v = (0, 1) (by condition (4.2)). Using Corollary 2.6 again, the directional derivative of Gx,y in direction v is nonnegative at any point of the segment [(p, q), (c, d)]. Therefore, by Lemma 2.7, we get that Gp,q (x, y) Gc,d (x, y). Case (iii). Then (c, d) ∈ H2 yields that c > 0 and d < 0. The inequality ad − bc 0 is obviously valid if (a, b) ∈ H1 , because then a and b are nonpositive. In the case (a, b) ∈ H2 ,
438
P. Czinder, Z. Páles / J. Math. Anal. Appl. 301 (2005) 427–438
we have that b < 0 < a, therefore, the condition (4.3) ensures that ad − bc 0 also holds. Thus the direction bc − ad (1, −1) (p − a, q − b) = c+d is co-directional with the vector u = (1, −1). Using Corollary 2.6, we can see that the directional derivative of Gx,y , in the direction u is nonnegative at any point of the segment [(a, b), (p, q)] ⊂ H1 ∪ H2 . Hence, by Lemma 2.7, it follows that Ga,b (x, y) Gp,q (x, y). Observe that (p, q) ∈ H2 , hence we have that the segment [(p, q), (c, d)] is also contained in H2 . On the other hand, (c + d − a − b)d c ,1 , (c − p, d − q) = c+d d
hence (c − p, d − q) is co-directional with the vector v = dc , 1 . In view of Corollary 2.6 again, we obtain that the directional derivative of Gx,y in direction v is nonnegative at any point of the segment [(p, q), (c, d)] ⊂ H2 . Therefore, by Lemma 2.7, we get that Gp,q (x, y) Gc,d (x, y). The Cases (iv), (v) are completely analogous to Cases (iii), (ii), respectively. Therefore, the details are left to the reader. 2
References [1] P. Czinder, Zs. Páles, A general Minkowski-type inequality for two variable Gini means, Publ. Math. Debrecen 57 (2000) 203–216. [2] P. Czinder, Zs. Páles, An extension of the Hermite–Hadamard inequality and an application for Gini and Stolarsky means, J. Inequal. Pure Appl. Math. 5 (2004), Article 42, 8 p. (electronic). [3] C. Gini, Di una formula compressiva delle medie, Metron 13 (1938) 3–22. [4] J. Hadamard, Étude sur les propriétés des fonctions entières et en particulier d’une fonction considérée par Riemann, J. Math. Pures Appl. 58 (1893) 171–215. [5] E. Neumann, J. Sándor, Inequalities involving Stolarsky and Gini means, Math. Pannon. 14 (2003) 29–44. [6] Zs. Páles, Inequalities for sums of powers, J. Math. Anal. Appl. 131 (1988) 265–270. [7] F. Qi, Generalized abstracted mean values, J. Inequal. Pure Appl. Math. 1 (2000), Article 4 (electronic).