E. R. LOVE TYPE LEFT FRACTIONAL INTEGRAL INEQUALITIES Текст научной статьи по специальности «Математика»

14

Probl. Anal. Issues Anal. Vol. 9(27), No3, 2020, pp. 14-30

DOI: 10.15393/j3.art.2020.8730

UDC 517.51

G. A. ÀNASTASSIQU

E. R. LOVE TYPE LEFT FRACTIONAL INTEGRAL INEQUALITIES

Abstract. Here first we derive a general reverse Minkowski integral inequality. Then motivated by the work of E. R. Love [4] on integral inequalities we produce general reverse and direct integral inequalities. We apply these to ordinary and left fractional integral inequalities. The last involve ordinary derivatives, left Riemann-Liouville fractional integrals, left Caputo fractional derivatives, and left generalized fractional derivatives. These inequalities are of Opial type.

Key words: Minkowski integral inequality, Opial inequality, Rie-mann-Liouville fractional integral, fractional derivatives.

2010 Mathematical Subject Classification: 26A33, 26D10, 26D15.

1. Introduction. This paper deals with ordinary and left fractional integral inequalities. We are motivated by the following results:

Theorem 1. (Hardy's Inequality, integral version [3, Theorem 327]) If f is a complex-valued function in Lr (0, to), || • || is the Lr (0, to) norm and r > 1, then

Theorem 2. [4] If s ^ r ^ 1, 0 ^ a < b ^ to, y is real, u (x) is decreasing and positive in (a,b), f (x) and H (x,y) are measurable and non-negative on (a,b), H (x,y) is homogeneous of degree -1,

x

(1)

0

x

© Petrozavodsk State University, 2020

and

then

where

f (x)r xY 1u (x) dx

\\Hf llr ^ C ||f lis

bt

C = H (l,t) t

xY (x)dx

dt.

Here a is to mean 0 if a b

b = 00.

0 or b = 00 or both; and bt is to mean 00 if

An application of Theorem 2 follows:

Theorem 3. [4] If p> 0, q> 0, p + q = r ^ 1, 0 ^ a <b ^ ro, y < r, u (x) is decreasing and positive in (a,b), f (x) is measurable and nonnegative on (a,b), Ia is the left Riemann-Liouville operator of fractional integration defined by

(Iaf ) (x)

(x -1)

a—1

Y (a)

-f (t) dt for a > 0,

10 f (x) = f (x), where r is the gamma function, and 13 f is defined similarly for ft ^ 0, then

[(Iaf)(x)]p[(I13f)(x)]qxY-ap-3q-1u(x)dx ^ C f (x)rxY-1u(x)dx, (7

where

C

r(i - r)

Y \ \P

T(i - r)

y \ \q

r (a + 1 - \r (ft + 1 - £

Also Theorem 2 implies Theorem 1 (by [4]), just take a = 0, b = ro, Y = 1, s = r > 1, u (x) = 1 and H (x,y) = X.

2. Main Results. We start with a general result, see also [2].

Theorem 4. (Reverse Minkowski integral inequality) Let (X, A, y) and (Y, B,v) be a-finite measure spaces and let 0 < p < 1. Here f is a

b

r

1

r s

x

b

b

nonnegative function on X x Y with f (x,y) > 0 for almost all x G X, almost all y G Y and J (f (x, y))p dv (y) < to for almost all x G X.

Y

Then

' f (x,y) dp (x) ) du (y) J > J y J (f (x,y))P (y)J dp(x),

Y X X Y

if left-hand side is finite.

Proof. Notice that J" f (x,y) dp (x) > 0, for almost all y G Y and

X

Y X

f (x, y) dß (x)) dv (y) > 0

We observe that

f (x,y) dß (x)) dv (y)

YX

p-i

f (x,y) dß (x)){J f ^^ y) dß (x>)

Y X X

f (x,y) [J f ^^ y) dß (x>)) dß (x)

Y X X

(by Tonelli's theorem)

p-i

dv (y) dv (y) =

f (x,y){ f (x',y) dß (x>) dv (y)

XY

X

dß (x) ^

(by applying the reverse Holder's inequality in the inside we get)

(f (x,y))pdv (y)

f(x',y)dß(x')) dv(y)

p-1

(9)

XY

YX

dß(x) (10)

(f (x,y))pdv (y)) p dß(x)

f(x,y)dß(x)) dv(y)

p-1

XY

YX

p

p

Finally, divide both ends of (10) by f (f f (x, y) dß (x)}Pdv (y)

Y X

to obtain (9). □

p-1 p

> 0

We continue with a reverse analog of Theorem 2. The proof involves a special kind of variation of reverse Minkowski integral inequality which we establish completely. We present

Theorem 5. Let 0 < r < 1, 0 < a < b < x>, f (x) and H (x,y) are measurable and non-negative on (a, b), (a, b)2, respectively, H (x, y) is homogeneous of degree -1,

x

(Hf )(x) = j H (x,y) f (y) dy, (11)

a

and suppose that

b

\\Hf 11^]= (/ (Hf )(x)r dx)1 < rc, (12)

and ||/ ||r[ab] is defined similarly.

We assume that H (l,t) / (x,t) > 0, for almost all t E [X, l], for almost all x E [a,b], and H^ ||/||p [a bt] < <x>, for almost all t E [|, l] . Then

i

||H/ ||P, [a 'b] ^ i ^^ ||/ ||P' a,bt]dt. (13)

J tr

a b

Proof. For a < x < b the homogeneity of degree — l of H gives

i i

(H/) (x) = j H (x,xt) / (xt) xdt = J H (1,t) / (xt) dt,

a a

x x

where t = X. As a ^ y ^ x, then 0 <t ^ l. We will prove first (0 < r < l)

b 1 i ||H/ ||p, [a, b] = (/( / H (l,t) / (xt) dt) P ck) r ^ (14)

1 b

>11 Î H (l,t)r f (xt)r dx)rdt = (*).

a a b t

Indeed we observe that

b 1 b 1

f H (1,t) f (xt) dt)rdx = I ( I x[a>1] (t) H (1,t) f (xt) dt) dx

a 0

(where x is the characteristic function) b 1 1

X[ x ,1] (t)H (l,t)f (xt)dtU X[ x ,1] (t')H (l,t')f (xt')dt') \dx

r-1

a0 b 1

"x[ X ,1] (t) H (1,t) f (xt)( i X[a ,1] (t) H (1,t!) f (xt1) dt) r-1dt\dx

a0

1 b

0

(by Tonelli's theorem) 1

X[ x ,1] (t) H (1,t)f (xt)( X[a ,1] (t ) H (1,t )f (xt')dt') dx\dt =

r1

1 b

H (1,t) f (xt) ^ Î H (1,1/) f (xt1) dtr 1dx

dt >

a a b t

(by reverse Holder's inequality applied inside)

1 b

b 1

H(l,i)/(xt)rdxj r( J [j H(1 ,t')f (xt')dt')rdx

dt = (15)

a a b t

1b b 1 r-r

f (H (1,t) / (xt))r dx) r dt)( f ( f H (1,t) / (xt) dt) r dx

a a b t

Clearly here, by the assumptions, it holds J ( J H (l,t) f (xt) dt) dx > 0 and

a V a /

x

all we did they make sense.

1

1

rr

r

(b 1 r ) r-1

Finally, we divide both ends of (15) by (f (f H (1 ,t)f (xt)dt)r dx\ r > 0, to

a a '

X

validate (14), which is a particular case of a reverse Minkowski type integral inequality.

By (14) we continue

1 b 1 bt (*) = / (j f (xt)r tdx)1 dt = j (j f (y)r dy)1 dt = (16)

1

rH (M)„„, dt

1 lu llr,l 11

proving (13). □

We give a reverse left fractional inequality.

Corollary 1. (to Theorem 5) Let 0 < r < p with r < l, 0 <a< b< <x, / is measurable and non-negative on (a, b) such that / (x) > 0 almost everywhere on [a, b]. Here Ia is the left fractional Riemann-Liouville integral operator of order a > 0, defined by

x 1

(ia/) (x)=j{x/ (t) dt, i0/ (x)=/ (x), (17)

a

V x E [a,b], and suppose that ||x"aIa/ (x) ||P[ab] < x>. We assume that [a bt] < for almost all t E [|, l]. Then

1

(Ia/(x))p(/(x))p-px~apdx > ||/ ||P' [a'bt]dt) P|/HPp~P.

a

b (18)

(x — y)a 1

Proof. In Theorem 5, let H (x, y) =-^ . . for x > y > a and a > 0,

v xaT(a) y

which is homogeneous of degree — l. Then (H/) (x) = x-aIa/ (x), and so

by (13)

1 1 a-1

l|Hf Hr,[a,b] > J llf Hr,[a,bt] dt =^1 (1 -f If lr,[a,bt] dt.

a a

b b

(19)

Here 0 < r < p, hence 0 < r < 1, also < 0, and f (x) > 0 almost

p ' r—p ' J \ /

everywhere in [a,b]. Next we apply the reverse Holder's inequality:

b

x—a (Iaf) (x))p (f (x))r—p dx >

(x—a (Iaf )(x)Y dx) (f (x))r dx)

1

(19) 1 f r(1- t)a—1 V

||Hf ||P,[a,b] |f |r—P 19 Yl^yj tt ||f ||r, [a,bt] dtj ||f |r

iP.[a.b]!f |r—P V . ? ||f Hrja.*]^ ||f H^ '

(20)

proving the claim. □

Next we present a reverse Opial type [5] inequality.

Corollary 2. (to Corollary 1) Let 0 < r < p with r < 1, m G N, 0 < a < b < to, f : [a,b] ^ R such that f(m—1) is an absolutely continuous

m)

is

function over [a, b], where f (a) = f (a) = ... = f(m 1) (a) = 0, and f( non-negative, with f(m) (x) > 0 almost everywhere over [a,b]. We assume that ||x—mf (x) ||r,[a,b] < to, and ||f(m) ||r,[a,bt] < to, a.e. for t G [a, 1]. Then

b

f (f (x))p (f(m) (x))r—p x—mpdx >

> (mhwU||f(m)|r,[a,bt]d^P|f(m)|r—p. (21)

^ m—1 \ p

((m - 1)\)p\J t~

a b

Proof. By Taylor's formula with integral remainder we have

x

r (x- t)m—

f (x) = J ({m_:1)l f(m) (t) dt = (Imf(m)) (x) , V x G [a, b] (22)

b

" — p

then apply Corollary 1 for f(m). □ We need

Definition 1. Let a > 0, n = [a] ([■] is the ceiling), f e ACn ([a,b]) (i.e. f(n-1) is absolutely continuous function). The left Caputo fractional derivative is given by

x

Daf (x) = v(l J (x - t)n-a-1 f(n) (t) dt (23)

V(n - a) J

a

and exists almost everywhere for x in [a,b], D0af = f, see [1, p. 394]. We mention

Corollary 3. [1, p. 395] Let a > 0, n = [a], f e ACn ([a,b]), and f(k) (a) = 0, k = 0,1,...,n - 1. Then

x

f (x) = Ya)i (x - t)a-1 Daf (t) dt = IaD"af (x), V x e [a, b]. (24)

a

We give a reverse left fractional Opial type [1] inequality.

Corollary 4. (to Corollary 1) Let 0 < r < p with r < 1, 0 < a < b < ro, a > 0, n = [a], f e ACn([a,b]), and f (k)(a) = 0, k = 0,1,... ,n - 1. Assume here that Daaf is non-negative over (a,b) such that D<aaf > 0 almost everywhere on [a, b]. Suppose that \\x-af (x) \\r,[a,b] < ro and \\D<aaf \\r [abt] < ro, for almost all t e [f ,1S. Then

(f (x))p (DU (x))r-p x-apdx >

i 1

^ (1 ~ f WDU llr,^] dt) \\D:j W- . (25)

Proof. By Corollaries 1, 3, see also (23). □ We need

b

Definition 2. [1, pp. 7-8] Let v > 0, n := [v] [•] the integral part, and a := v — n (0 < a < 1); x, x0 G [a,b] C R such that x > x0, x0 is fixed. Let f G C ([a,b]) and define

x

(JX0 f) (x) := J (x — t)v—1 f (t) dt, xo ^ x ^ b, (26)

xo

the left Riemann-Liouville integral. We define the subspace CX0 ([a,b]) of Cn ([a,b]):

CXo ([a,b]) := f G Cn ([a,b]) : J—^ f G C1 ([xo,b])} . (27)

For f G CX0 ([a,b]) we define the left generalized v-fractional derivative of f over [x0, b] as

DXo+f := DJX— af(n) (f(n) := Dnf). (28)

Notice DX0+f G C ([xo,b]). We also need

Theorem 6. [1, from Theorem 2.1, p. 8] Let f G CX0 ([a,b]), x0 G [a,b] fixed.

1) If v > 1, and f(i) (x0) = 0, i = 0,1,...,n — 1, then f (x) = (JX°DX0+f (x), all x G [a,b] : x ^ x0.

2) If 0 <v< 1, then f (x) = J0 DX+ f) (x), all x G [a,b] : x ^ x0.

That is, in both cases we have

X

f (x) = rjv)J(x — t)v—l (DX0+f) (t) dt, x0 ^ x ^ b. (29)

X0

If x0 = a, we get

X

f (x) = Yty) j (x — t)—1 (D+f) (t) dt = (JDV+f) (x), all a ^ x ^ b.

a

We give another reverse left fractional Opial type inequality.

Corollary 1. (to Corollary 1) Let 0 < r < p with r < 1, 0 < a < b < ro, v > 0, n = [v]; f e Cva ([a,b]), such that f(i) (a) = 0, i = 0,1,...,n — 1 for only the case of v ^ 1. Assume here that D^+f is non-negative over (a,b) such that D%+f > 0 almost everywhere on [a,b]. Suppose that l\x-v f (x) ||r,[a,b] < ro and \\Dva+f \\r,[a,bt] < ro, for almost all t e [f,1]. Then

b

j (f (x))p (Da+ f (x))r-p x-vpdx >

a

1 _ 1

s tw(I f" llwt)p\\D:+f \\;:-p. (31)

a b

Proof. By Corollary 1, Theorem 6, see also (28). □

We need the following representation result.

Theorem 7. [1, p. 395] Let v ^ y + 1, Y ^ 0. Call n = \v|, m := \y| . Assume f e ACn([a, b]), such that f(a) = 0, k = 0,1,... ,n — 1, and DVaf e L^ (a, b). Then DJaf e C ([a, b]), DJaf (x) = Im-f(m) (x), and

x

DYaf (X) = r — y) J (X — t)V-Y-1 DVaf (t) dt = (I V-YDVaf) (x) , (32)

a

V x e [a,b].

Remark 1. (to Theorem 7) By Corollary 3 we also have

x

f (x) = Vv)j(x — t)V-1 DVaf (t) dt = (IVDVaf) (x) , (33)

a

V x e [a,b].

It follows left fractional direct Opial type integral inequalities.

Theorem 8. If p > 0, q > 0, p + q = r ^ 1, 0 ^ a < b < ro, y < r, w (x) is decreasing and positive in (a,b). Let v ^ y + 1, Y ^ 0, call n = \v|, f e ACn ([a, b]) : f(k) (a) = 0, k = 0,1,...,n — 1; D U e L^ (a, b), with DVaf ^ 0 over (a,b). Then

b

J (f (x))p (DYaf (x))q xY-vp-(v-Y)q-1w (x) dx ^

b

^ C i (Df )(x))r xY—1u (x) dx, (34)

where

r (1 — r) w^C1 — r)

r (v +1 — lr (v —

C= ( TT^^ T^—. (35)

rr

Proof. Directly from Theorem 3. Notice that b

J [(IVDIJ) (x)]p [(F-YDf (x)] xY—vp—(v—Y)q—1u (x) dx ((32)=(33))

a

b

= J (f (x))p (DYaf (x))q xY—vp—(V—Y)q—1 U (x) dx. (36)

a

So, in applying (7), now instead of f we take D*Vaf. □

Theorem 9. If p > 0, q > 0, p + q = r > 1, 0 ^ a < b < to, y < r, u (x) is decreasing and positive in (a,b). Let v ^ Y + 1, Y ^ 0, i = 1, 2, call n = \v], f G ACn ([a,b]) : f(k) (a) = 0, k = 0,1,...,n — 1; D*af G L^ (a, b), with D*f > 0 over (a,b). Then

b

J (DUf (x))p (DY2 f (x))q xY—(v—YI)p—(v—Y2)q—1u (x) dx ^ a b

^ C* J ((DVaf )(x))r xY—1u (x) dx, (37)

a

where

C * =, ^ — ?) A V ^ — r) (38)

Proof. Use of Theorem 7 and similar to Theorem 8. □ Corollary 1. All as in Theorem 8. Then b

J (DYaf (x))P ((DVaf) (x))q xY—(V—Y)p—1u (x) dx ^

q

q

b

< C i ((DU)(x))r x1-lu (x) dx, (39)

where

r (l - Y)

C 'r (v - y + 1 - r Proof. By Theorems 3, 7. □ We need

Remark 2. [1, p. 26] Let v ^ y + 1, Y ^ 0, n = [v], x0 G [a, b] fixed, f G C£0 ([a,b]) : f(i) (xo ) = 0, i = 0,1,...,n - 1. Then

x

(DJo+f) (x) = Yyih^) I (x - t)( V-Y)-1 (DVo+f) (t) dt, (40)

x0

which is continuous in x on [x0, b]. We continue with

Theorem 10. If p > 0, q > 0, p + q = r ^ 1, 0 ^ a < b < to, y < r, u (x) is decreasing and positive in (a,b). Let v ^ Y + 1, Y ^ 0, n = [v],

f G Cv ([a,b]) : f W (a) = 0, i = 0, l,...,n - i. Assume that Dva+f ^ 0 on (a,b). Then

b

J (DY+f (x))P (DV+f (x))q xY-(v-Y)p-1^ (x) dx ^

b

Cij (DV+f (x))r xY-1u (x) dx, (41)

a

r (i - Y)

where

C = ( r{1 - ?) C1 Vr (v - y + 1 - r

Proof. By Theorem 3 and see Remark 2. □ We make

Remark 3. By [4], see Theorem 2, let s = r ^ 1, 0 ^ a < b ^ to,

Y is real, u (x) is decreasing and positive in (a,b), f (x) and Hk (x,y)

p

(k = 1,... ,n) are measurable and non-negative on (a,b), Hk (x, y) is homogeneous of degree —1,

(Hk f )(x)= Hk (x,y) f (y) dy, k = 1,...,n, (42)

and

' 1

r

f (x)r xY-1w (x) dx) , (43)

then where

Hk fll ^ Ckl Ifl, (44)

Ck = Hk (1,t) t-r dt, k = 1,...,n. (45)

Here a means 0 if a = 0 or b = oo or both; and bt means oo if b = oo.

b

n

Let now pk > 0 such that pk = r.

k=1

We notice the following (apply generalized Holder's inequality) b „ b

n n f c \-rk

J] (Hk f (x))pk xY-1 w (x) dx ^ n ( / (Hk f (x))r xY-1w (x) dx 1

U— 1 U— 1 \ J '

k=1 k=1

n (44) n / n \ E Pk

n llHkf l\Pk 44 n CPpk ||f l\Pk = n Cf If lir=1 = (46) k=1 k=1 ^k=1 '

b

II CPk) llf l\r = C^j f (x)r xY-1W (x) dx),

k=1 a

where

n

C := n CPpk. (47)

k

k=1

We have proved

b

r

1

Theorem 11. Let 0 ^ a < b ^ ro, y is real, w (x) is decreasing and positive in (a,b), f (x) and Hk (x,y) (k = 1,... ,n) are measurable and non-negative on (a,b), Hk (x,y) is homogeneous of degree —1,

x

(Hk f) (x) = J Hk (x,y) f (y) dy, k =1,..., n. (48)

a

n

Let pk > 0^ pk = r ^ 1. Then k=1

b b f n (Hkf (x))Pk xY-1w (x) dx ^ C^J f (x)r xY-1w (x) dx^J ,

49)

a k=1

where

n / 1

~ ~ Hk (l,t) t-Ydt) . (50)

n / 1 \pk _ Y

r I

C=n(/

fc=iv a

Next we give an application.

n

Theorem 12. Here pk > 0^ pk = r ^ 1. Let 0 ^ a < b ^ j < r,

k=1

u (x) is decreasing and positive in (a,b), f (x) is measurable and nonnegative on (a,b), Iak is the left Riemann-Liouville operator of fractional integration defined by

x

nx- t)ak-1

(Iak f) (x) = J ( J f (t) dt, for ak > 0, (51)

a

and

Iak f (x) = f (x) ,for ak = 0; k =1,...,n.

Then

b n n b

f n ((Iak f) (x))pk ak Pk-1 u (x) dx ^ C^f f (x)r x'-1 u (x) dx^J ,

a k=1 a

(52)

where

5 = S( fS) " (53)

Proof. Here we apply Theorem 11. Inequality (52) derives from (49) directly. We let

(x — y)a; 1

Hk (x,y) =-——— for x > y > a and ak > 0.

xak r (ak)

Then Hk f (x) = x-ak Ia; f (x), k G{1,...,n} . Notice that i i -1 J Hk (1,t) t-rdt = j (1-(t^j t-rdt ^

< f (1 -^-11-rdt = r (1 - ^ 7), (54)

r (ak) r (ak + 1 - r' ,

for k G {1,..., n} .

Therefore ( ( )

C = II ( r (1 -

1\ \ p;

r ,

\\ r (ak + 1 - r) □

Next we give general left fractional direct Opial type integral inequalities.

N

Theorem 13. Here pj > 0 : Pj = r > 1. Let 0 ^ a < b < to,

j=1

Y < r, u (x) is decreasing and positive in (a,b). Let v ^ Yjj + 1, Yjj ^ 0, j = 2,..., N, n = \v], f G ACn ([a,b]) : f(k) (a) = 0, k = 0,1,...,n - 1, and Dlaf G L^ (a, b), with DVaf ^ 0 over (a, b). Then

b N N _

/. . / — \ Pj 7—vpi—Y"" (V—7,')p, — 1

(f (x))Pl n (DYaf (x)) j x j j u (x) dx ^

j=2

a

N

r{1 - r) v1 -Q^ r{1 - r) y,

r (v + 1 - rv f=wr (v - Yj + 1 - r

X

b

x( I (D Vaf (x))r x7-1u (x) dx). (55)

Proof. By Theorem 12, use of Theorem 7 and (33). □ Theorem 14. All as in Theorem 13. Then

b N N _

/. . / — \ p, 7— E (v—7,)p, — 1

(D Vaf (x))P^ (D?a f (x)) x j=2 u (x) dx ^

j=2

a

^ (n ( r (v - % Y ) )Pj )(/ (D Vaf (x))r x7—1u (x) dx). (56)

j=2 r a

Proof. By Theorem 12, use of Theorem 7. □

Corollary 1. All as in Theorem 13, and = Y3 = ... = Yx, YA+1 = Yx+2 = ... = %, and Y^+I = Y^+2 = ... = Yn . Then

b A M N

i (DVaf (x))Pl (DYAf (x))SPj (D>f (x))^1 Pj (DYN f (x))j^+1 Pj x

A \ / M \ / N

( V_YA) I _ Pj ) - . _ _

xx Vj=2 y Vj=M+i / ^ (x) dx ^

■Ym) ( _

Vj=A+1

A

\ E Pj

j=2

Y )

r ) /

N

E Pj ,

1 =M+1 /

r (i - Y) jPj f r (i - r) ^,=_+iPj.

r (v - yx + i - rv vr (v - % +i - r).

X ( w ^ , Y ) V^ Pj ( i (D Vaf (x))r xY-1u (x) dx). (57)

r (v - yn + 1 - r)

a

Proof. By Theorem 14. □ We finish with

N

Theorem 15. Here pj > 0 : Pj = r > 1. Let 0 ^ a < b < to,

j=1

Y < r, u (x) is decreasing and positive in (a,b). Let v ^ % + 1, % ^ 0,

j = 2,...,N,n = [v], f G Cva ([a,b]) : f(k) (a) = 0, k = 0,1,...,n - 1. f

a+ f

a

Assume that DV+f > 0 on (a,b). Then

b N N _

p y __\ P Y_ "y y ( v_Y ' )P _1

(DV+f (x))P1 n (Da7+ f (x)) j x j=2 j j w (x) dx ^ j=2

^ (n (r (v+ Y) T)(/ (DV (x))r(x) d*y (58)

r a

Proof. By Theorem 12, use of (40). □

Comment. With the exhibited methods above one can derive all kinds of variation of left fractional Opial type integral inequalities, as well as of ordinary differentiation ones, due to lack of space we omit this task.

References

[1] Anastassiou G. A. Fractional Differentiation Inequalities. Springer, Heidelberg, New York, 2009.

DOI: https://doi.org/10.1007/978-0-387-98128-4

[2] Benaissa B. On the reverse Minkowski's integral inequality. Kragujevac Journal of Mathematics, 2022, vol. 46(3), pp. 407-416.

[3] Hardy G. H., Littlewood J. E., Polya G.Inequalities.Cambridge, UK, 1934.

[4] Love E. R. Inequalities like Opial's inequality. Rocznik naukowo dydakty-czny WSP w Krakowie, Pr. Mat., 1985, vol. 97, pp. 109-118.

[5] Opial Z., Sur une inégalité. Ann. Polon. Math., 1960, vol. 8, pp. 29-32. DOI: https://doi .org/10.4064/ap-8-1-29-32

Received May 31, 2020. In revised form,, June 09, 2020. Accepted July 02, 2020. Published online July 12, 2020.

George A. Anastassiou Department of Mathematical Sciences University of Memphis Memphis, TN 38152, U.S.A. E-mail: [email protected]