Functional Inequalities Involving Numerical Differentiation Formulas of Order Two
Functional Inequalities Involving Numerical Differentiation Formulas of Order Two
Tomasz Szostok 0
0 Institute of Mathematics, University of Silesia , Bankowa 14, 40007 Katowice , Poland
We write expressions connected with numerical differentiation formulas of order 2 in the form of Stieltjes integral, then we use Ohlin lemma and LevinStechkin theorem to study inequalities connected with these expressions. In particular, we present a new proof of the inequality f f Communicated by Mohammad Sal Moslehian.
HermiteHadamard inequality; Differentiation formulas; Convex functions

f (t )dt
1 Introduction
Writing the celebrated Hermite–Hadamard inequality
f (t )dt ≤
B Tomasz Szostok
in the form
F (y) − F (x )
y − x
y − x
Remark 1 Let f, F, : [x , y] → R be such that = F, F = f, let ni , mi ∈
N ∪ {0}, i = 1, 2, 3; ai, j ∈ R, αi, j , βi, j ∈ [0, 1], αi, j + βi. j = 1, i = 1, 2, 3; j =
1, . . . , ni ; bi, j ∈ R, γi, j , δi, j ∈ [0, 1], γi, j + δi, j = 1, i = 1, 2, 3; j = 1, . . . , mi . If
the inequality
i=1
in=21 a2,i F (α2,i x + β2,i y)
y − x
in=31 a3,i (α3,i x + β3,i y)
im=21 b2,i F (γ2,i x + δ2,i y)
(y − x )2
y − x
i=1
im=31 b3,i (γ3,i x + δ3,i y)
(y − x )2
The simplest expression used to approximate the secondorder derivative of f is of
the form
Remark 2 From numerical analysis it is known that
This means that for convex g and for G such that G
G(x ) − 2G
= g we have
+ G(y)
Lemma 1 (Ohlin [7]) Let X1, X2 be two random variables such that EX1 = EX2 and
let F1, F2 be their distribution functions. If F1, F2 satisfy for some x0 the following
inequalities
E f (X1) ≤ E f (X2)
for all continuous and convex functions f : R → R.
f (x )d F1(x ) ≤
f (x )d F2(x )
for all continuous and convex functions f : [a, b] → R it is necessary and sufficient
that F1 and F2 verify the following three conditions:
F1(t )dt ≤
F2(t )dt, x ∈ (a, b),
F1(t )dt =
F2(t )dt.
Let now f : [x , y] → R be any function and let F, : [x , y] → R be such that F = f
and = F. We need to write the expression
(x ) − 2
in the form
where F1 : [0, 1] → R is given by
(0) − 2
(1) =
f d F1,
F1(t ) :=
−2x 2 + 4x − 1
Proof Let F : [0, 1] → R be such that
enough to do the following calculations
f d F1 =
4x f (x )dx +
(−4x + 4) f (x )dx = 2F
− 0 · F (0)
4F (x )dx
4F (x )dx − 0 · F (1) − 2F
= 4 (1) − 8
Remark 4 Observe that if
equality is satisfied
and f are such as in Proposition 1 then the following
(x ) − 2
ds dt.
After this observation it turns out that inequalities involving the expression (9)
were considered in the paper of Dragomir [3] where (among others) the following
inequalities were obtained
f (t )dt.
2 The Symmetric Case
We start with the following remark.
f (t )dt
F∗(t )dt =
Ta f (x , y) =
if a ≤ 0, then
if a ≤ 2, then
if a ≥ 6, then
if a ≥ −6, then
f (t )dt,
f (t )dt,
Ta f (x , y) ≤ f
Ta f (x , y) ≤
≤ Ta f (x , y),
Furthermore,
if a ∈ (2, 6), then the expressions Ta f (x , y), f x+y are not comparable in the
2
class of convex functions,
if a < −6, then the expressions Ta f (x , y), f (x)+ f (y) are not comparable in the
2
class of convex functions.
F1(t ) :=
f d F1 =
f d F1 =
+ 1
2
a
= 1 − 2 (F (1) − F (0)) + 2a
dt = F
− F
(0) − 2
F (t )2adt
ϕ(s) :=
F3(t ) − F1(t )dt, s ∈ [0, 1]
f d F1 ≤
f d F2,
F3(t ) :=
F4(t )dt
This theorem provides us with a full description of inequalities which may be obtained
using Stieltjes integral with respect to a function of the form (17). Some of the obtained
inequalities are already known. For example from (12) and (13) we obtain the
inequality
(y − x )2
f (t )dt,
ds dt.
(y − x )2 x
f (t )dt ≤ 3
y − x x
(y − x )2 x
y − x x
(y − x )2 x
(y − x )2 x
(y − x )2 x
y − x x
In Corollary 1 we obtained inequalities for the triples:
(y − x )2
(y − x )2
f (t )dt,
f (t )dt, f
(y − x )2
Remark 7 Using the functions: F1 defined by (10) and F5 given by
F5(t ) :=
t = 0,
t ∈ 0, 21
t ∈ 21 , 1
t = 1,
we can see that
1 1
+ 6 f (y) ≥ (y − x )2
3 The Nonsymmetric Case
Now, in contrast to the symmetric case (Remark 5), it is possible to prove
inequalities using just one quadratic function but before we do this we shall present a
nonsymmetric version of Hermite–Hadamard inequality involving only the
primitive function of f.
Proposition 2 Let x , y be some real numbers such that x < y and let α ∈ [0, 1].
Let f : [x , y] → R, be a convex function, let F : [x , y] → R be such that F = f. If
Sα1 f (x , y) is defined by
y − x
then the following inequality is satisfied:
Proof As usually, the proof will be done on the interval [0, 1]. Define the functions
F6, F7, F8 : [0, 1] → R by the following formulas:
F6(t ) :=
F8(t ) :=
F6(t )dt =
F7(t )dt =
f d F6 = f (1 − α),
f d F7 = α f (0) + (1 − α) f (1)
f d F8 = Sα1 f (0, 1).
(4 − 6α)F (y) + (2 − 6α)F (x )
y − x
(6 − 12α)( (y) −
(y − x )2
then the following conditions hold true:
if α ∈ 31 , 23 then
Sα2 f (x , y) ≤ α f (x ) + (1 − α) f (y),
Sα2 f (x , y) ≥ f (αx + (1 − α)y),
Sα2 f (x , y) ≤ Sα1 f (x , y)
and if α ∈ 31 , 21 ∪ 21 , 23 then Sα1 f (x , y) and Sα2 f (x , y) are incomparable in the
class of convex functions.
Proof Take
and let F6, F7, F8 be defined so as in Proposition 2. Then we have
f d F1 =
(6 − 12α)t + 6α − 2 f (t )dt
(1) −
F9(t )dt =
F8(t )dt =
F7(t )dt =
F6(t )dt.
f d F9 ≤
f d F7,
Thus for every convex function f we have
f d F6 ≤
f d F9,
F9(t )dt =
F6(t )dt
f d F6,
4 Concluding Remarks and Examples
F1(t ) :=
ax 2 + (1 − α)x
cx 2 + (1 − cα − c)x + cα
where c = − 1−αα , must be used. Since the description of all possible cases in
Theorem 2 was already quite complicated, we shall not present these inequalities in
details here.
F1(t ) :=
t ≤ 41 ,
f (t )dt,
8 (x ) − 16
− 16
(y − x )2
where f : [x , y] → R is any convex function and
= f.
Remark 10 We have
t 2d F6(t ) = (1 − α)2.
This means that for two values of α : 3−6√3 and 3+6√3 we have 01 t 2d F9(t ) =
01 t 2d F6(t ). Moreover, as it was mentioned in the proof of Theorem 3, the
functions F9, F6 have in this case two crossing points. This implies (see [2,9]) that the
inequalities:
3 −
√3
√3
3 −
are satisfied by all 2convex functions f : [x , y] → R
Remark 11 It is easy to see that all inequalities obtained in this paper in fact
characterize convex functions (or 2convex functions). This is a consequence of results
contained in paper [1].
1. Bessenyei , M. , Páles, Zs: Characterization of higher order monotonicity via integral inequalities . Proc. R. Soc. Edinb. Sect. A 140 ( 4 ), 723  736 ( 2010 )
2. Denuit , M. , Lefevre , C. , Shaked , M. : The sconvex orders among real random variables, with applications . Math. Inequal. Appl. 1 ( 4 ), 585  613 ( 1998 )
3. Dragomir , S.S. : Two mappings in connection to Hadamard's inequalities . J. Math. Anal. Appl. 1(167) , 49  56 ( 1992 )
4. Dragomir , S.S. , Gomm, I.: Some new bounds for two mappings related to the HermiteHadamard inequality for convex functions . Numer. Algebra Control Optim . 2 ( 2 ), 271  278 ( 2012 )
5. Levin , V.I. , Stecˇkin , S.B. : Inequalities. Am. Math. Soc. Transl . 14 ( 2 ), 1  29 ( 1960 )
6. Niculescu , C.P. , Persson , L.E. : Convex Functions and Their Applications . A Contemporary Approach. CMS Books in Mathematics/Ouvrages de Mathématiques de la SMC , vol. 23 . Springer, New York ( 2006 )
7. Ohlin , J. : On a class of measures of dispersion with application to optimal reinsurance . ASTIN Bull . 5 , 249  266 ( 1969 )
8. Olbrys´ , A. , Szostok , T.: Inequalities of the HermiteHadamard type involving numerical differentiation formulas . Results Math . 67 , 403  416 ( 2015 )
9. Rajba , T.: On the Ohlin lemma for HermiteHadamardFejer type inequalities . Math. Inequal. Appl . 17 ( 2 ), 557  571 ( 2014 )
10. Szostok , T.: Ohlin's lemma and some inequalities of the HermiteHadamard type . Aequ. Math. 89 , 915  926 ( 2015 )
11. Szostok , T.: Levin Steckin Theorem and Inequalities of the HermiteHadamard Type. (submitted) arXiv:1411 .7708 [math.CA]