CC BY
3166 Probl. Anal. Issues Anal. Vol. 9 (27), No 3, 2020, pp. 66-82
DOI: 10.15393/j3.art.2020.8270
UDC 517.518.86, 517.218.244, 517.927.2
B. Bayraktar
SOME NEW GENERALIZATIONS OF HADAMARD-TYPE MIDPOINT INEQUALITIES INVOLVING FRACTIONAL INTEGRALS
Abstract. In this study, we formulate the identity and obtain some generalized inequalities of the Hermite-Hadamard type by using fractional Riemann-Liouville integrals for functions whose absolute values of the second derivatives are convex. The results are obtained by uniformly dividing a segment [a,b] into n equal sub-intervals. Using this approach, the absolute error of a Midpoint inequality is shown to decrease approximately n2 times. A dependency between accuracy of the absolute error (e) of the upper limit of the Hadamard inequality and the number (n) of lower intervals is obtained.
Key words: convexity, Hadamard inequality, Holder's inequality, Power-mean inequality, Riemann-Liouville fractional integrals. 2010 Mathematical Subject Classification: 26A51, 26D15
1. Introduction. The well-known definition of a convex function in literature (for example, [4] and references therein) is
Definition 1. The function g : [a, b] ^ R is said to be convex, if
g № +(1 - t) Z) ^ tg (£) + (1 - t) g (Z) (1)
for all £,Z E [a, b] and t E [0,1].
It is well known that the double Hermite-Hadamard-type inequality plays a very important role in nonlinear analysis. This inequality is stated as follows [8]:
Theorem 1. Let g : I C R ^ R be a convex function defined on an interval I and let a,b e I, with a < b; then the following double inequality holds:
© Petrozavodsk State University, 2020
< ^) < nkj*™* ^
a
The theory of fractional integro-differential calculus, as an extension of classical analysis, plays an important role in pure and applied science. A. M. Nakhushev describes in the monograph [14] the fundamental elements and qualitatively new properties of fractional calculus operators. Moreover, their application to solving problems of mathematical modeling of various complex systems and processes (in living and non-living systems) has a fractal structure. Butkovsky et al. [3] presented the history of the development of fractional calculus and a detailed analysis of the problems of using fractional integro-differential calculus to describe dynamics of various systems and control processes. Examples of physical systems in need of the fractional analysis theory are presented.
In the literature, there are various definitions of the fractional integral (for example, see [13], [15]), but the Riemann-Liouville definition is among the most widely used in many applications of fractional calculus.
Definition 2. [11] Let g G L1 [a,b]. The Riemann-Liouville integrals Jaa+ g and J- g of order a > 0 are defined, respectively, by
1
r^a)
Jaa+g = (C - t)a-1 g (t) dt, x > a
and
Jb-g (0 = ^ i (t - 0a-1 g (t) dt, x <b
r (a) J ?
oo
where r (a) = f e-uua-1du is the well known Gamma function. Here, for
a = 0, J°a+ g (0°= J0-g (0 = g (C) and for a = 1
b
Jl+ g (C) = Jl- g (C) = J g (0
a
The vast majority of studies on the theory of integral inequalities use two classical inequalities. This is the Holder's inequality and its other form: the power-mean inequality:
b
Theorem 2. (The Holder inequality [12]). Let p> 1 and p + 1 = 1. If u(i) and v(i) are real functions, defined on [a,b], and \u\p, \v\q e L[a,b], then:
b b b j \u (i) v(£)\ di ^ (j \u a)\p di)1 (/\v(o\q di)q
a a a
This inequality turns into equality if and only if A\g(i) \p = B\g(i) \q almost everywhere; A and B are constants.
Theorem 3. (Power mean inequality [12]). Let q ^ 1 and 1 + 1 = 1. If u(i) and v(i) are real functions, defined on [a, b], and if \u\p, \v\q e L[a, b], then:
b b b i \u (i) v(i)\ di ^ ( i \u (i)\ di)1-1 ( i \u (i)\\v(i)\q di
In the recent years, many authors (see [1], [5-7], [10-13], [15] and references therein) have studied Hermite-Hadamard type inequalities involving fractional integrals for improvements and generalizations. In these papers, new inequalities for functions from various convexity classes were obtained. For example, in [5], [9], [16], some new Hadamard-type integral inequalities were obtained for functions whose first derivatives are convex functions and take values at the intermediate points of the integration interval.
In this study, using the identity formulated for fractional integrals, we obtain some new generalizations of the Hadamard type inequalities for functions whose absolute values of the second derivatives are convex. In addition, the obtained results clarify the errors of the quadrature formula for the numerical integration.
2. Main Results. Let n e N and a,b e R and let a < b, n ^ 1. The interval [a,b] with a uniform step h = b--a is divided into n subintervals
n
[a, b] = U [ifc-i,ifc ], where & = a + ih, i = 0,1, 2,...,n. k= 1
Lemma 1. Let g : I C R ^ R be a twice differentiable function on I°; I ° is the interior of I .If g e L [a,b], where a,b e I, then Va > 1 the following equality holds:
^h-T1 £ [Ja-1g(ik-1) + j;+"1g(ik)] - £ g (^) k=1 k k k=1
2a—2 h2 n
(hk + I2k), (4)
where
a
k=i
h = —, C = a + ih, i = 0,1, 2,...n, 9k = + n 2
0.5
4-a J'
Ilk = tag" ((1 - t) Ck—i + tCk) dt,
0
i
I2k = (1 - t)a g" ((1 - t) Ck—i + tCk) dt.
0.5
Proof. Integrate the integrals under the sum operator twice by parts and take into account that h = Ck — Ck-1 to get
0.5
4-a J'
Iik = tag" ((1 - t) Ck—i + tCk) dt
0
0.5
1a
g' (Ok) - r I ta—ig' ((1 -1) Ck—i + tCk) dt =
2 h k h
0
0.5
g' ^) ag ^) + I ta—2g ((1 - t) Ck—i + tCk) dt.
2a h 2a—ih2 h2
0
Similarly, for the second integral: i
I2k = i (1 - t)a g'' ((1 - t) Ck—i + tCk) dt
0.5
i
= - 21hg' (Ok ) + h/ (1 - t)a—i g' ((1 -1) Ck—i + tCk ) dt
0.5
i
g' (Ok) ag (°k) , a(a - 1) fn .,a~2
2a h 2a—ih2 h2
0.5
+ ~t2 (1 - t)a—2g ((1 -1) Ck—i + tCk ) dt.
By summing these integrals, we get:
a m \ a (a — 1)
I1k + hk = — ^0=2^2 9 (Ok) + -h2—x
x
0.5
' ta=2g ((1 — t) £k-i + t^k) dt+l (1 — t)a-2 g ((1 — t) £k-i + dt
0.5
If we make a change of variables (1 — t) £k-1 + t£k = x in the last equality, we get
a m \ a (a — 1) hk + 12k = — -¿0=2^ 9 (°k) +--X
Qh a—2 ^k a-2
x (x — ^ ~ 9 (x) dx W^^I^Lix) dx
. J ha-1 J ha-1
?fc-1 Qh
a a (a — 1)r(a — 1)
>9(Ok) + ~v~ :r " J<a=19(ik-i) + jX^^)
2«~2h2^ k> • ha+1
Q- z^-1' '
Multiply both sides of the last equality by [^—¡i^) and take into account the Gamma-function property:
2a~2h2 ln. 2a-2r(a)
(hk + hk) = —9 (Ok) + ha_\ J x
1k 2k — k
a h
1 1
x
ja=19dk-1)+jy9(<ik) Qh Qh
(5)
By summing both sides of (5) over k, we obtain (4). The proof is complete. □
Remark. If we choose n = 1, then from (4) we get the equality in [1] (equality (2) for m =1).
Theorem 4. Let 9 : I C R ^ R be twice differentiable function on I°. If 9" e L[a,b], where a,b G I and \9"\ is a convex function; then the following inequality holds Va > 1:
2a-2r(a) n n
ha-1 ^ Jq- 9(^k-1) 1 jQ
%-1) + J+- 9(^k
h
k=1 k=1
E \Ja-:19(Zk-1)+j;t-19(^k) — ^ 9(Ok)
h
2n
* 80(0+1) £< 1 ' + ' 9"(&) \ ) • (6)
where h = ^, & = a + ih, i = 0,1, 2,...,n and 0k = Sfe-12+Sfc. Proof. From Lemma 1, by using the triangle inequality, we obtain
2a-2r(a) n
h«P- £ [Ji"1g(ik-i) + Ja+1g(ik)
k=1
2«—2h2 n
Eg (*k) € £ (\I1k\ + \I2k\) . (7) k=1
k=1
Since the \g"\, the function is convex; using inequality (1) for the first integral, we get
\Iik \
1/2
€
J tag ((1 -1) ik—1 + tik) dt 0
1/2 1/2
€ \g"(ik—1)/ ta(1 - t)dt + \g"(ik)\ j ta+ldt. 00
Calculating these integrals, we get
a j 3 1
\I1k\ € (a + 1Ha + 2) 2«+2 \g"(ik—^ + (a + 2) 2«+ \g"(ik)\. (8)
Similarly, for the second integral, we write
\I2k\ = J (1 - t)ag"((1 - t)ik—1 + tik)dt € 1/2
1 j 3
€ (0+2)2^ \g"(ik—+ (a + 1Ha + 2)2«+2 'g"(ik)l ■ (9)
By summing the relevant sides of inequalities (8) and (9), we get:
\I1k \ + \ 12k \ € (a + 11) 2a+! [\g//(ik—1)\ + \g"(ik )\] ■ (10)
By multiplying both sides of the inequality (10) by the expression 2 ^ and by summing over k, we get
2a—2h2 ™ h2
¿d IV V ^ 1 I I 1 T 1 \ 10
% II —." / I
ik—1)
a
E (\I1k\ + \I2k\) € 8a (a + 1)g [\g"(ik—1) \ + \g"(ik)\] ■ (11)
1
By taking (7) and (11) into account, we obtain (6). The proof is completed. □
Corollary 1. If we choose a = 2, then from (6), we get the inequality
ba
g(x)dx - A (yi,y2,...,Vn)
(b - a)2 „ ,
€ W"A (z1,z2,
,zn) , (12)
where
A(.,.) - arithmetic mean of n real numbers,
yk =g (Ok), zk = \g"(ik—1)\ + \g"(ik )\, k = 1,2,...,n.
Proof. For a = 2, for the left-hand side of the inequality (6), we get
r(2)
n [ ] n
h £ J1-g(ik—1) + Jkg(ik)1 - £ g (Ok) k=1
n
ba
?fc
£
k=1
k=1
?fc
J g(x)dx + J g(x)dx Sfc-l Qt
n1
£g(Ok)
k=0
n
ba
k=1
g(x)dx) - Y^ g(Ok)
k=1
n
ba
g(x)dx g(Ok)
k=1
(13)
For the right-hand side of the inequality (6), we write h2
8a(a
£(\g//(ik—1)\ + \g"(ik )\)
k=1
it [(\g//(ik—1)\ + \g//(ik )\)] ■ (14)
k=1
By taking into account the equalities (13) and (14) from (6), we obtain
n
ba
n (b-n)2 n
g(x)dx - £ g (Ok) € £ [(\g//(ik—1)\ + \g//(ik)\)] ■
k=1
k=1
b
1
b
1
b
By dividing both sides of the last inequality by n, we obtain (12). The proof is complete. □
Remark. It is easy to verify that if we choose n =1 and a = 2, then, from (12), we get the inequality
¿-J«* - №
< [|g'(-)| +
:i5)
This inequality was obtained by M. Sarikaya et al. (see [17, Proposition 1]) for convex functions.
Proposition 1. If the absolute error £ is given, then the following inequality holds for n:
n >
b - - h
6£
where l|g"ll = SUPxe[a,b] \g//(x)\ ■
Proof. Indeed, from the right-hand side of the inequality (14), we have
^ £ [is" (&-0! + !/ (&)!] < (b4=ni)2 (2n y ») = ^ »9
k=1 b-«)2 II „n
48n3
24n-
:i6)
Since ^—«r ||g"» ^ £, we get n ^
b-a. /Vl
- V 6e
. The proof is complete. □
Remark. If ||g"|| = supx€[ab] |g"(x)|, then it follows from inequality (15) that
b
1
b-
g(x)dx — g
- + b 2
^ ||g"|
^ 24 "y |
and from (12), taken into account inequality (16), we get
1 f 1
g(x)dx--V g (dk)
b-
n
k=1
^ (b — -)- ||g"| ^ 24n- »g |
These two inequalities show that when the interval is divided into n sub-intervals, the error of an Hadamard-type inequality decreases n2 times.
b
Remark. Let R(g,h) be the error estimation of the Midpoint rule of numerical integration; then from (12) we obtain the well-known error estimate [2]:
R(g, h) = ^h2 ||g"|| .
Theorem 5. Let g : I С R ^ R be a twice differentiable function on I°. If g" G L [a,b], where a,b G I and | g"\q is a convex function, then, for all а > 1, q ^ 1 and t G (0,1), the following inequality holds:
2a Г(а)
h
a—1
k=1
Y.\ja-1g(^-i) + Ja-1g(ik) -£ g (Ok)
h2
k=1
8а (а + 1)
£ Fk, (17)
where
h
Fk
ba
n
, = a + ih, i = 0,1, 2,... ,n, Ok
k=1
Ck-1 + Ck 2 ;
(а + 1) |g//(gk-1)|q + (а + 3) |g"(Ck)|
2 (а + 2)
+
+
(а + 3) |g"(£k-1)|q + (а + 1) |g"(Ck)|q
2 (a + 2)
Proof. From Lemma 1, we obtain, by using the triangle inequality,
2a-2Г(а)
n
n
ha-1 T, [ja-1g(Ck-1) + ja-1g(Ck)J - £ g (Ok) k=1 k=1
<
n1
2a-2h2
< -£ (|l1k| + 112k|). (18)
а
k=0
By using the well-known Power-mean integral inequality (3) and since \ 9"\q is a convex function, we get
|l1k |
1/2
tag" ((1 -1) Ck-1 + tCk) dt
1/2
^ I J tadt\ x 0
1-1
1/2
1/2 _ i
|g"(Ck-1)|^ ta (1 - t) dt + |g"(Ckta+1dt 4 00
q
q
(a + 1) 2a+1
x
i-1
x
(a + 1) (a + 2) 2a+2 (a + 3) |g''(£k-i)|q + (a + 1) |g''&)|q
Simplifying the last statement, we get:
Ihk| ^
Note that
1
(a + 1) 2a+1
(a + 3) lg"(Çk-i)|q + (a + 1) |g'(&)|q 2 (a + 2)
:i9)
|l2k |
(1 - t)a g"((1 - t)Ck-i + tCk )dt
1/2
1/2
tag"(tik-i + (1 - t)ik)dt
so, similarly to the first integral, we can write down an inequality for I2k:
112k | ^
1
(a + 1) 2a+1
(a + 1) W'^k-i)|q + (a + 3) \g"(ik)|q
2 (a + 2)
Summing the relevant sides of the inequalities (19) and (20), we get
1
(20)
|1ik | + |l2k | ^
(a + 1) 2a+1
Fk.
2a - 2h2
Multiplying both sides of the last inequality by the expression summing over k, and taking into account (18), we obtain (17). The proof is completed. □
Corollary 1. Taking a = 2, we obtain, from (17):
b — a
g(x)dx - A (yi,V2,.. .,Vn)
<
(b - a)2 48n2
A(Fi,F2,...,Fn), (21)
where
A(.,.) — arithmetic mean of n real numbers,
1
1
q
1
q
b
1
yk =g (Ok), Ok = ^k ^k , k = 1, 2,... ,n,
2
Fk
3 |g"(£k)|9 + 5 |g"(Ck-i)|q 8
+
5 |g//(Ck )|9 + 3 |g//(Ck-i)|q 8
Here we get (12) for q =1.
Remark. Taking a = 2 and n = 1, we get, from (17):
bhfg(x)dx - )
(b — a)2 ^
^ --- x F1,
^ 48 1
where
Fi
3 |g»|9 + 5 |g//(b)|q 8
1/9
+
5 |g//(a)|q + 3 |g//(b) |9
8
1/9
This inequality for convex functions has been obtained by M. Sarikaya, N. Aktan, and the author (see [17, Proposition 5]).
Theorem 6. Let g : I C R ^ R be a twice differentiable function on I°. If g// G L [a, b], where a,b G I and | g//|9 is a convex function, then, for all a,q and p > 1, such that 1 + 1 = 1, the following inequality holds:
2a-2r(a)
h
a—1
£ [^(^-1) + J^gtek)] - £ g (Ok) k=1 k=1
h2 ( 1
16a \aq — q + 3
£ Wk, (22)
k=1
where
h =-, = a + ih, i = 0,1, 2,...,n, Ok = 1 + ^k,
n2
Wk = [# |g//(Ck-1)|9 + |g//(Ck)|9]1 + [|g//(Ck-1 )|9 + # |g"(£k)|9] ^ = qa — q + 4
qa - q + 2
Proof. From Lemma 1, we obtain, using the triangle inequality:
q
q
b
q
q
2a-2r(a)
k=1
ha[j:-;19(tk-1) + j;-19(^k)\ — £ 9 (°k) ■ - k=1
2a-2h2
*
a
£(\l1k\ + \l2k\). (23)
k=1
Using the well-known Holder integral inequality (2) and since \ 9"\q is a convex function, we get
\l1k \
1/2
ta9" ((1 — t) ik-1 + tCk) dt
1/2
tpt1 ta-19" ((1 — t) Ck-1 + ttk) dt
*
1/2
*
p )pdt
1/2 1/2
\9//(Ck-1)\^ tqa-q+1(1—t)dt+\9"(ik)\qj tqa-q+2dt 00
3
1\ 3
1
(qa — q + 3)2qa-q+3 _
qa—^ i9"Kk-1)iq+\9//( & )\q
Given the fact that
3
1\ p
1
(qa — q + 3)2qa-q+3 _
q 1
1
2a+2 \qa — q + 3
we can write an inequality for I1k:
\ hk \ *
1
1
2«+M qa — q + 3
Since
9// (£k-1) \q + \ 9"(tk) \q ]1 . (24)
\ I2k \
(1 — t)a 9// ((1 — t) Ck-1 + tCk) dt
1/2
1/2
ta9// (tCk-1 + (1 — t) Ck) dt
p
q
2
q
1
we can write an inequality for I2k, similarly to the first integral:
II2kI ^ ^ (qa—¿+3) 1 [lg//(€k-1)|9 + # \g"(ik)|9]1. (25) Summing the relevant sides of the inequalities (24) and (25), we get
\I1k\ + \I2k\ ^ (-1-19 ■ Wk;
\ 1k\t\ 2k\ ^ 2a+^qa — q + 3J k'
by multiplying both sides of the last inequality by the expression 2a a2fe2, summing over k, and taking into account (23), we obtain (22). The proof is completed. □
Corollary. Note that 1 = 1 — 1. Choose a = 2; then we get from (22):
b
g(x)dx — A (y1,y2,...,yra)
ba
^ ^ (q) A(W1,W2 ,...,Wn), (26)
where
A(.,.) — arithmetic mean of n real numbers,
1
yk =g (Ok), Ok = ^k-1n+ ^k , k = 1, 2,...,n, ^ (q) = 1
Wk
2 ^ Vq + 3
q+4
\g// (Ck-1)\9 + \g//(Ck )\9
i _ _ i
q
q
+
\g//(Ck-1 )\9 + ^ \g//(Ck )\9
q+2
Here i1 ^ ^ (q) < 1 for all q > 1, since lim ^ (q) = 1 and lim ^ (q) = 1.
16 9^1+ 4 9^+ro
For q ^ 1+, we get (15) from (26).
3. Some Applications To The Special Means. Let us consider some special means for two real numbers a and 3.
1) Arithmetic mean : A(a,3) = ;
2) Quadratic mean : Q(a,3) = \Ja2 + 32;
3) Logarithmic mean : L(a,3) = ln, a,3 > 0 and a = 3;
4) Generalized logarithmric mean :
(a, if a = 3,
(n + 1)(3 - a), ifa = 3,n G
Now, by using the results, we will give some applications to special means of positive real numbers.
Proposition 2. Let g(x) = Vl + x2; then Vx G [0,b] the following inequality holds:
A [Q (b, 1), 1 ln (2 A (b, Q (b, 1))) ] - ^A (yi,y2 ,...,yn)
<
* ^A [1,Q-3(b, 1),2n4A(t1,t2,...,tn-1)] ,
where
yk = Q [(2k — 1) b, 2n] ,k =1, 2,...,n,
tm = Q-3 (m2b2,n2) , m = 1, 2,... ,n — 1,
n is the number of subintervals of the interval [0, b].
Proof. This inequality follows from Corollary 1.
Indeed, for the left-hand side of inequality (12) with a = 0, we obtain:
b b ^__ ^ ^_
-- 9(x)dx = - V1 + x2dx =----1----—-- =
b a b 2 2b
Q(b-1) + ln(2A (b-Q <b-1))) = AÎQ (b, 1), 1 ln [2A (b, Q (b, 1))] '
2 2 b
as C = a + i ^ = i n ,i = 0,1, 2,... ,n,
n n' 7 7 7 7 7
yk = 3[Ck-i + M = 1 + f Ck-i + M 2 V(2k - 1)2b2 + 4n2
2 2 2n
2nQ [(2k - 1)b, 2n]
As g"(x)
(i+x2)3/2
on the right-hand side (12), we get:
b2 b2 n A(zi,z2,...,zn) = J]
48n2
48n3
b2
48n3
b2
48n3
b2 16n3
n—l
k=i
n
1
L (€2—1 + 1)3/2 + (€2 1
+
1 + 2 k=iV (k2b2 + n2)3/V ' (b2 + 1)3/2
n— l
1 + Q—3(b, 1) + 2n3 ^ Q-3 (kb,n)
k=i
A
1, Q (b, 1), 2n4A(ti,t2,...,tn—i)
Here zfc = |g"(£fc— i)| + |g"(&)| and tk = Q—3 (kb,n). □
Proposition 3. Let g(x) = ex and let the interval [a, b] be divided into n subintervals; then Vx G [a,b], the following inequality holds:
\L (ea, eb) - e"A (yi, y2,..., yn) \ ^ EA (ti,t2,... ,tn )
where
24n2
yk = e(2k—1)h, h = —, k =1, 2,
n
E = «•{A[A 1 (35)'A 1 (5^ I
tm = e(m—i)h, m =1, 2,... ,n.
n,
Proof. This inequality follows from Corollary 1. Indeed, for the left-hand side of the inequality (21) we get:
1 f 1 f
g(x)dx = -- I exdx
ba
ba
eb - e" ba
L(e",eb)
and, since ^ = a + ih, h = b——", i = 0,1, 2,... ,n,
yk = g (^) = e"+(2k+i) h
i
1
b
b
Since g"(x) = ex, we can write for the right-hand side of the inequality (21),
Fk
k =
3 |g'(&)|q + 5 |g"(6k-i)r 8
3e(a+kh)q + 5e(a+(k-1)h)q
+
8
+
5 lg"(jk)|q + 3 |g"(ek-i)|q 8
5e(a+kh)q + 3^e(a+(k-1)h)q 8
= e(a+(k-1)h)
3ehq + 5" q 5ehq + 3"
+
_ 8 _ _ 8 _
3ehq + 5" q 5ehq + 3"
+
_ 8 _ _ 8 _
e
(k-1)h
2ea
A
AU -eqh,- },A-eqh, -
5
5 qh 3
e
(k-1)h
4 '4J1 V4 4, Thus, A (F1,F2,...,Fn) = 2ea ■ E ■ A (t1,t2,... ,tn). □
References
[1] Bayraktar B. Some integral inequalities Of Hermite-Hadamard type for differentiable (s,,m)-convex functions via fractional integrals., TWMS J. App. Eng. Math,. 2020, vol. 10(3), pp. 625-637.
[2] Boglaev Yu. P. Computational Mathematics and Programming. Moscow, Higher School, 1990.
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DOI: https://doi.org/10.1134/S0005117913040012
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Received March 26, 2020.
In revised form, June 18, 2020.
Accepted June 23, 2020.
Published online July 11, 2020.
Bursa Uludag University,
Faculty of Education, Gorukle Campus,
16059, Bursa, Turkey.
E-mail: [email protected]
