Volume 8 (2007), Issue 2, Article 46, 5 pp.
SUPERQUADRACITY OF FUNCTIONS AND REARRANGEMENTS OF SETS
SHOSHANA ABRAMOVICH
D EPARTMENT OF M ATHEMATICS
U NIVERSITY OF H AIFA
H AIFA , 31905, I SRAEL
abramos@math.haifa.ac.il
Received 26 December, 2006; accepted 29 May, 2007
Communicated by S.S. Dragomir
A BSTRACT. In this paper we establish upper bounds of
n X
|xi − xi+1 |
xi + xi+1
+f
,
f
2
2
i=1
xn+1 = x1
when the function f is superquadratic and the set (x) = (x1 , . . . , xn ) is given except its arrangement.
Key words and phrases: Superquadratic functions, Convex functions, Jensen’s inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
We start with the definitions and results of [1] and [5] which we use in this paper.
Definition 1.1. The sets (y− ) = y1− , . . . , yn− and (− y) = (− y1 , . . . ,− yn ) are symmetrically
decreasing rearrangements of an ordered set (y) = (y1 , . . . , yn ) of n real numbers, if
y1− ≤ yn− ≤ y2− ≤ · · · ≤ y −n+2
[ 2 ]
(1.1)
and
(1.2)
−
yn ≤
−
y1 ≤
−
yn−1 ≤ · · · ≤
−
y[ n+1 ] .
2
A circular rearrangement of an ordered set (y) = (y1 , . . . , yn ) is a cyclic rearrangement of (y)
or a cyclic rearrangement followed by inversion.
Definition 1.2. An ordered set (y) = (y1 , . . . , yn ) of n real numbers is arranged in circular
symmetric order if one of its circular rearrangements is symmetrically decreasing.
Theorem A ([1]). Let F (u, v) be a symmetric function defined for α ≤ u, v ≤ β for which
∂ 2 F (u,v)
≥ 0.
∂u∂v
001-07
2
S HOSHANA A BRAMOVICH
Let the set (y) = (y1 , . . . , yn ), α ≤ yi ≤ β, i = 1, . . . , n be given except its arrangement.
Then
n
X
F (yi , yi+1 ) ,
(yn+1 = y1 )
i=1
is maximal if (y) is arranged in circular symmetrical order.
Definition 1.3 ([5]). A function f , defined on an interval I = [0, L] or [0, ∞) is superquadratic,
if for each x in I, there exists a real number C (x) such that
f (y) − f (x) ≥ C (x) (y − x) + f (|y − x|)
for all y ∈ I.
A function is subquadratic if −f is superquadratic.
Lemma A ([5]). Let f be a superquadratic function with C (x) as in Definition 1.3.
(i) Then f (0) ≤ 0.
(ii) If f (0) = f 0 (0) = 0, then C (x) = f 0 (x) whenever f is differentiable.
(iii) If f ≥ 0, then f is convex and f (0) = f 0 (0) = 0.
The following lemma presents a Jensen’s type inequality for superquadratic functions.
Lemma
P B ([6, Lemma 2.3]). Suppose
Pthat f is superquadratic. Let xr ≥ 0, 1 ≤ r ≤ n and let
x = nr=1 λr xr , where λr ≥ 0, and nr=1 λr = 1. Then
n
n
X
X
λr f (xr ) ≥ f (x) +
λr f (|xr − x|) .
r=1
r=1
If f (x) is subquadratic, the reverse inequality holds.
From Lemma B we get an immediate result which we state in the following lemma.
Lemma C. Let f (x) be superquadratic on [0, L] and let x, y ∈ [0, L], 0 ≤ λ ≤ 1, then
λf (x) + (1 − λ) f (y)
≥ f (λx + (1 − λ) y) + λf ((1 − λ) |y − x|) + (1 − λ) f (λ |y − x|)
t−1 X
≥ f (λx + (1 − λ) y) +
f 2λ (1 − λ) |1 − 2λ|k |x − y|
k=0
+ λf (1 − λ) |1 − 2λ|t |x − y| + (1 − λ) f λ |1 − 2λ|t |x − y| .
If f is positive superquadratic we get that:
t−1 X
λf (x) + (1 − λ) f (y) ≥ f (λx + (1 − λ) y) +
f 2λ (1 − λ) |1 − 2λ|k |x − y|
k=0
More results related to superquadracity were discussed in [2] to [6].
In this paper we refine the results in [7] by showing that for positive superquadratic functions
we get better bounds than in [7].
Theorem B ([7, Thm. 1.2]). If f is a convex function and x1 , x2 , . . . , xn lie in its domain, then
n
X
x1 + · · · + xn
f (xi ) − f
n
i=1
n−1
x1 + x2
xn−1 + xn
xn + x1
≥
f
+ ··· + f
+f
.
n
2
2
2
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 46, 5 pp.
http://jipam.vu.edu.au/
S UPERQUADRACITY AND R EARRANGEMENTS
3
Theorem C ([7, Thm. 1.4]). If f is a convex function and a1 , . . . , an lie in its domain, then
(n − 1) [f (b1 ) + · · · + f (bn )] ≤ n [f (a1 ) + · · · + f (an ) − f (a)] ,
where a =
a1 +···+an
n
and bi =
na−ai
,
n−1
i = 1, . . . , n.
2. T HE M AIN R ESULTS
Theorem 2.1. Let f (x) be a superquadratic function on [0, L] . Then for xi ∈ [0, L] , i =
1, . . . , n, where xn+1 = x1 ,
n
n−1X
(2.1)
f
n i=1
n
X
xi + xi+1
!
2
i=1
≤
+f
n
X
|xi − xi+1 |
2
i=1
n
X
!
−f
f (xi )
!!
i=1
n
X
xi
i=1
!
n
n
1X
−
f
n i=1
!
n
X
x
j
x i −
n
j=1
holds. If f 000 (x) ≥ 0 too, then
n
(2.2)
n−1X
f
n i=1
n
X
xi + xi+1
!
n
X
|xi − xi+1 |
!!
+f
2
2
i=1
n
x
bi + x
bi+1
n−1X
|b
xi − x
bi+1 |
f
+f
≤
n i=1
2
2
!
!
!
n
n
n
n
X
X
X
X
xi
1
xj ≤
f (xi ) − f
−
f xi −
,
n
n
n
i=1
i=1
i=1
i=1
j=1
where (b
x) = (b
x1 , . . . , x
bn ) is a circular symmetrical rearrangement of (x) = (x1 , . . . , xn ) .
Example 2.1. The functions
f (x) = xn ,
n ≥ 2,
x ≥ 0,
and the function
f (x) =
x2 log x, x > 0,
0,
x=0
are superquadratic with an increasing second derivative and therefore (2.2) holds for these
functions.
Proof. Let f be a superquadratic function on [0, L] . Then by Lemma B we get for 0 ≤ α ≤
1, 1 ≤ k ≤ n and xi ∈ [0, L] , xn+1 = x1 ,
(2.3)
n
X
i=1
n
n
n−k X
kX
f (xi ) =
f (xi ) +
f (xi )
n i=1
n i=1
n
n
n−k X
kX
=
(αf (xi ) + (1 − α) f (xi+1 )) +
f (xi )
n i=1
n i=1
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 46, 5 pp.
http://jipam.vu.edu.au/
4
S HOSHANA A BRAMOVICH
n
n−k X
≥
f (αxi + (1 − α) xi+1 )
n i=1
n
n−k X
(αf ((1 − α) |xi+1 − xi |) + (1 − α) f (α |xi+1 − xi |))
n i=1
!
Pn
Pn
X
n
x
x
1
i
i=1 i
+k f
+
f xi − i=1 .
n
n
n
i=1
+
For k = 1 and α = 21 we get that (2.1) holds.
2 F (u,v)
If f 000 (x) ≥ 0, then ∂ ∂u∂v
≥ 0, where
F (u, v) = f (u + v) + f (|u − v|) ,
Therefore according to Theorem A, the sum
n
X
xi + xi+1
|xi + xi+1 |
f
+f
,
2
2
i=1
u, v ∈ [0, L] .
xn+1 = x1 ,
is maximal for (b
x) = (b
x1 , . . . , x
bn ) , which is the circular symmetric rearrangement of (x) .
Therefore in this case (2.2) holds as well.
Remark 2.2. For a positive superquadratic function f , which according to Lemma A is also a
convex function, (2.1) is a refinement of Theorem B.
If f 000 (x) ≥ 0, (2.2) is a refinement of Theorem B as well.
Remark 2.3. Theorem B is refined by
Pn
n
X
n−1
i=1 xi
f (xi ) − f
≥
n
n
i=1
!
n
X
x
bi + x
bi+1
f
2
i=1
n
n−1X
xi + xi+1
f
,
≥
n i=1
2
because a convex function f satisfies the conditions of Theorem A for F (u, v) = f (u + v) .
The following inequality is a refinement of Theorem C for a positive superquadratic function
f , which is therefore also convex. The inequality results easily from Lemma B and the identity
!
n
n
n
X
X
1 X
f (ai ) =
f (aj ) (1 − δij )
n
−
1
i=1
i=1
j=1
(where δij = 1 for i = j and δij = 0 for i 6= j), therefore the proof is omitted.
Theorem 2.4. Let f be a superquadratic function on [0, L] , and let xi ∈ [0, L], i = 1, . . . , n.
Then
!
!
n
n
X
X
n
f (xi ) − f (x) −
f (yi )
n−1
i=1
i=1
!
n X
n
n
X
1
1 X
≥
f (|yi − xj |) (1 − δij ) +
f (|x − xi |) ,
n − 1 i=1 j=1
n − 1 i=1
P
i
, i = 1, . . . , n.
where x = ni=1 xni , yi = nx−x
n−1
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 46, 5 pp.
http://jipam.vu.edu.au/
S UPERQUADRACITY AND R EARRANGEMENTS
5
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[5] S. ABRAMOVICH, G. JAMESON
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AND
G. SINNAMON, Refining Jensen’s inequality, Bull. Sci.
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J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 46, 5 pp.
http://jipam.vu.edu.au/