Fractional calculus, in which the introduction of several distinct fractional integral operators is used in solving integeral inequalities, has proved useful in applications in such fields as physics, engineering, and computer science. The operators introduced include the Riemann-Liouville, Hadamard, Katugompola, and proportional fractional integral operators. The proportional fractional integral operator is particularly noteworthy in being a generalized fractional operator. Specific applications of generalized operators can be found, for example, in [8], and [6].

In [7], the weighted fractional integral is defined as follows. For an integrable function f on the interval [a,b] and for a differentiable function μ such that μ'(t)≠0 for all t∈[a,b], it is

 a+Iwβf(x)=1w(x)Γ(β)∫axμ′(s)(μ(x)−μ(s))β−1w(s)f(s)ds, x>a,

where w is a weighted (positive measurable) function.

Researchers have expanded and improved Minkowski's inverse inequality by applying it to fractional integral operators. This has resulted in the development of new mathematical tools that have enhanced our ability to solve problems in various fields. See [9], [10], [13].

In [1], the author presented the following generalization of the reverse Minkowski's inequality, for any measurable functions f,g>0 on (a,b) and p≥1, if 0<c<m≤αf(x)g(x)≤M for all x∈[a,b], then

M+αα(M−c)∫ab(αf(x)−cg(x))pdx1p≤∫ab fp (x)dx1p+∫ab gp (x)dx1p≤m+αα(m−c)∫ab (αf(x)−cg(x))p dx1p.

Using c=1 we get Sroysang's inequalities [12, Theorem 2.2] and if we put α=c=1 we obtain the Sulaiman's inequalities [11, Theorem 3.1]. In [2], the authors provide a generalization of the reverse Hölder's inequality. For λ,γ>0 and f,g,w>0 measurable functions on (a,b) and p>1 ,(1p+1p'=1), if 0<m≤αfλ(x)gγ(x)≤M for all x∈[a,b] we have

∫abfλ(x)w(x)dx1p∫abgγ (x)w(x)dx1 p′≤ Mm 1pq∫abf λ p (x)gγp′ (x)w(x)dx.

Moreover, a new version to the reverse Hölder's inequality with two parameters was has been presented on time scales in [3].

Motivated by the above literature, the present paper introduces a new definition of weighted fractional operators of a function with respect to another function. In Section 2, we present new versions of the reverse Minkowski-type inequality using k-weighted fractional operators with two parameters. Section 3 establishes the reverse Hölder-type inequality in fractional calculus using k-weighted fractional operators with two parameters. At the end of the paper, we conclude with a brief summary of the findings.

Let [a,b]⊆(0,+∞), where a<b. In this section, we present a definition of the k-weighted fractional integrals of a function f with respect to the function μ and we prove that they are bounded in a specified space.

Definition 2.1. Let β>0, k>0 and μ be a positive, strictly increasing differentiable function such that μ'(s)≠0 for all s∈[a,b]. The left and right sided k-weighted fractional integral of a function f with respect to the function μ on [a,b] are defined respectively as follows.

(2.1) a+ℑwμf(x)=1w(x)kΓk(β)∫axμ′(s)(μ(x)−μ(s))βk−1w(s)f(s)ds, x>a.
(2.2) b−ℑwμf(x)=1w(x)kΓk(β)∫xbμ′(s)(μ(s)−μ(x))βk−1w(s)f(s)ds, x<b,

where w is a weighted function and the k-gamma function defined by

Γk(β)=∫0∞tβ−1e−tkkdt.

When f(s)=1, we denote

 a+ℑwμ1(x)=1w(x)kΓk(β)∫axμ′(s)(μ(x)−μ(s))βk−1w(s)ds, x>a,

and

 b−ℑwμ1(x)=1w(x)kΓk(β)∫xbμ′(s)(μ(s)−μ(x))βk−1w(s)ds, x<b.

Remark 2.1. Let c>0 be a positive constant if f(s)=c, we get

 a+ℑwμc.(x)=c  a+ℑwμ1(x).

The space LpW[a,b] of all real-valued Lebesgue measurable functions f on [a,b] with norm conditions:

fpW=∫ab∣f(x)∣pW(x)dx1p<∞, 1≤p<+∞.

is known as weighted Lebesgue space, where W be a weight function ( measurable and positive ).

• Put W≡1, the space LpW[a,b] reduces to the classical space Lp[a,b].

• Choose W(x)=wp(x)μ'(x), we get

(2.3)LXwp[a,b]=f: fXwp=∫ab∣w(x)f(x)∣pμ′(x)dx1p<∞.

In the following Theorem we show that the k-weighted fractional integrals are clearly defined.

Theorem 2.1. The fractional integrals (2.1), (2.2) are defined for all functions f∈LXw1[a,b] and we have

(2.4) a+ℑwμf(x)∈LXw1[a,b],   b−ℑwμf(x)∈LXw1[a,b].

Moreover

(2.5) a+ℑwμf(x)Xw1≤Cf(x)Xw1,    b−ℑwμf(x)Xw1≤Cf(x)Xw1,

where

C=μ(b)−μ(a)βkΓk(β+k).

Proof. For all βk>0, by using Fubini's Theorem, we get

 a+ℑwμf(x)Xw1=∫abw(x)∣a+ ℑwμf(x)∣μ′(x)dx≤1kΓk (β)∫ab ∫ ax w(s)f(s) μ′(s)(μ(x)−μ(s))βk−1μ′(x)dsdx=1kΓk(β)∫abw(s)f(s) ∫s b (μ(x)−μ(s)) β k −1μ′(x)dxμ′(s)ds=1βΓk(β)∫abw(s)f(s)μ(b)−μ(s)βk μ ′(s)ds≤ μ(b)−μ(a) βk Γk(β+k)∫abw(s)f(s) μ ′(s)ds=C f(x) Xw1 .

Similarly

∫abw(x)∣b−ℑwμf(x)∣μ′(x)dx≤μ(b)−μ(a)βkΓk(β+k)f(x)Xw1.

This gives us our desired formulas (2.5) and (2.4).

Setting μ(τ)=τ, then  a+Iwμf(x) and  b-Iwμf(x) reduce to the k-weighted fractional integral of Riemann-Liouville operator of order β≥0.

RL1kf(x)=1w(x)kΓk(β)∫ax(x−s)βk−1w(s)f(s)ds, x>a.
RL2kf(x)=1w(x)kΓk(β)∫xb(s−x)βk−1w(s)f(s)ds, x<b,.

Setting μ(τ)=lnτ, then  a+Iwμf(x) and  b-Iwμf(x) reduce to the k-weighted fractional integral of Hadamard operator of order β≥0.

H1f(x)=1kΓk(β)∫axlnxsβk−1f(s)dss, x>a>1.
H2f(x)=1kΓk(β)∫xblnsxβk−1f(s)dss, x<b.

Setting μ(τ)=τρ+1ρ+1 where ρ>0, then  a+Iwμf(x) and  b-Iwμf(x) reduce to the k-weighted fractional integral of Katugompola operator of order β≥0.

K1f(x)=(ρ+1)1−βkw(x)kΓk(β)∫axxρ+1−sρ+1βk−1w(s)sρf(s)ds, x>a.
K2f(x)=(ρ+1)1−βkw(x)kΓk(β)∫xbsρ+1−xρ+1βk−1w(s)sρf(s)ds, x<b,.

Setting μ(τ)=(τ-a)θθ for θ>0 (respectively μ(τ)=(b-τ)θθ ), then  a+Iwμf(x) (respectively  b-Iwμf(x) ) are reduce to the k-weighted fractional integral of conformable operator of order β≥0.

C1f(x)=θ1−βkw(x)kΓk(β)∫ax(x−a)θ−(s−a)θβk−1w(s)f(s)(s−a)1−θds, x>a.
C2f(x)=θ1−βkw(x)kΓk(β)∫xb(b−x)θ−(b−s)θβk−1w(s)f(s)(b−s)1−θds, x<b,.

For example see [5]. The most important feature of the k-weighted fractional integrals  a+Iwμf(x) and  b-Iwμf(x) is that they give certain types of the k-weighted fractional depends on the choice of the function μ.

We present the following Lemma [4], [3], that is used to prove our results.

Lemma 2.1. Let 1<q≤p<∞ and f,W be non-negative measurable functions on [a,b]. We suppose that 0<∈tabfr(s)W(s)ds<∞ for r>1, then

(2.6)∫ab fq(s)W(s)ds≤∫abW(s)dsp−qp ∫ab f p (s)W(s)dsqp

and

∫ab f q′(s)W(s)ds≥∫abW (s)ds p′−q′ p′ ∫ab fp′ (s)W(s)dsq′p′ .

Proof. If p=q we get equality and for p≠q using the Hölder's integral inequality with pq>1, we have

∫abfq(s)W(s)ds=∫ab Wp−qp (s)fq(s)Wqp(s)ds≤ ∫ a b W(s)dsp−qp ∫ a b f p (s)W(s)dsqp.

Since 1<q≤p<∞⟹1<p'≤q'<∞, thus the proof of the second inequality is similar to the first one.

Corollary 2.1. Let 1<q≤p<∞, f be non-negative measurable function on [a,x] and μ be a positive strictly increasing differentiable on [a,b] and  a+Iwμ is the operator defined by (2.1), then

(2.7) a+ℑwμfq(x)1q≤ a+ℑwμ1(x)p−qpq  a+ℑwμfp(x)1p,

and

 b−ℑwμfq(x)1q≤ b−ℑwμ1(x)p−qpq  b−ℑwμfp(x)1p.

Proof. Using the inequality (2.6) by taking W(s)=1w(x)kΓk(β)μ'(s)(μ(x)-μ(s))βk-1w(s), we obtain that

∫0x1 w(x)kΓk(β)μ′(s) (μ(x)−μ(s)) βk −1w(s)fq(s)ds≤ ∫ 0 x 1 w(x)k Γ k (β) μ ′ (s) (μ(x)−μ(s)) β k −1w(s)ds p−qp×  ∫ 0 x 1 w(x)k Γ k (β) μ ′ (s) (μ(x)−μ(s)) β k −1 w(s) f p (s)ds qp,

this gives the desired results.

Let 0≤a<b<+∞, w be a weight function and f,g be a positive measurable functions on [a,b], suppose that fp, gp∈LXw1[a,b], where

(3.1)LXw1[a,b]=f: fXw1=∫ab∣w(x)f(x)∣μ′(x)dx<∞.

Theorem 3.1. Let f,g>0,1≤q≤p<+∞,α>0 and

(3.2)0<c<m≤αf(s)g(s)≤M,  for all s∈[a,x],

then

(3.3)M+αα(M−c)  a+ℑwμ (αf(x)−cg(x))qdx1q≤  a+ℑwμ fq(x)dx1q+ a+ℑwμ1(x) p−qq   a+ℑwμ gp(x)dx1p≤m+αα(m−c)   a+ ℑwμ1(x) p−qq   a+ℑwμ (αf(x)−cg(x))pdx1p,

and

(3.4)M+αα(M−c)  b−ℑwμ (αf(x)−cg(x))qdx1q≤  b−ℑwμ fq(x)dx1q+   b− ℑwμ1(x) p−qq   b−ℑwμ gp(x)dx1p≤m+αα(m−c)   b− ℑwμ1(x) p−qq   b−ℑwμ (αf(x)−cg(x))pdx1p.

Proof. From the assumption (3.2) we get

0<1c-1m≤1c-g(s)αf(s)≤1c-1M,

hence

cMM-c≤cαf(s)αf(s)-cg(s)≤cmm-c,

that yields

Mα(M-c)(αf(s)-cg(s))≤f(s)≤mα(m-c)(αf(s)-cg(s)),

taking the qth power of the above inequality and multiplying by the positive quotient

1w(x)kΓk(β)μ′(s)(μ(x)−μ(s))βk−1w(s)

we obtain

Mα(M−c)q1w(x)kΓk(β)μ′(s)(μ(x)−μ(s))βk−1w(s)(αf(s)−cg(s))q≤1w(x)kΓk(β)μ′(s)(μ(x)−μ(s))βk−1w(s)fq(s)≤mα(m−c)q1w(x)kΓk(β)μ′(s)(μ(x)−μ(s))βk−1w(s)(αf(s)−cg(s))q,

integrating with respect to s over [a,x], we get

(3.5)Mα(M−c)  a+ℑwμ (αf(x)−cg(x))qdx1q≤  a+ℑwμ fq(x)dx1q≤mα(m−c)  a+ℑwμ (αf(x)−cg(x))qdx1q,

applying the inequality (2.7) on the right-hand side of (3.5), we get

(3.6)Mα(M−c)  a+ℑwμ (αf(x)−cg(x))qdx1q≤  a+ℑwμ fq(x)dx1q≤mα(m−c)  a+ℑwμ1(x)p−qpq   a+ℑwμ (αf(x)−cg(x))pdx1p.

Using the assumption (3.2), we obtain

0<m-c≤αf(s)-cg(s)g(s)≤M-c,

therefore

αf(s)-cg(s)M-c≤g(s)≤αf(s)-cg(s)m-c,

for p≥1 we deduce that

1M-cpαf(s)-cg(s)p≤gp(s)≤1m-cpαf(s)-cg(s)p,

multiplying by the positive quotient 1w(x)kΓk(β)μ'(s)(μ(x)-μ(s))βk-1w(s) and integrating with respect to s over [a,x], thus

1M−c  a+ℑwμ (αf(x)−cg(x))p1p≤  a+ℑwμ gp(x)1p≤1m−c  a+ℑwμ (αf(x)−cg(x))p1p.

therefore

(3.7)1M−c a+ℑwμ1(x)p−qpq a+ℑwμ (αf(x)−cg(x))p 1p≤ a+ℑwμ1(x)p−qpq a+ℑwμ gp (x)1p≤1m−c a+ ℑwμ1(x)p−qpq a+ℑwμ (αf(x)−cg(x))p 1p.

Applying the inequality (2.7) on the left-hand side of (3.7), we get

(3.8)1M−c  a+ℑwμ (αf(x)−cg(x))q1q≤ a+ℑwμ1(x)p−qpq  a+ℑwμ gp(x)1p≤1m-c a+Iwμ1(x)p-qpq a+Iwμ(αf(x)-cg(x))p1p.

Adding the inequalities (3.6) and (3.8), we get the required inequality (3.3). The proof of the inequality (3.4) is similar to the proof of the inequality (3.3).

We present some results which are special cases of Minkowski's reverse type inequalities via the k-weighted fractional integral (2.1) with two-parameters in the Corollaries mentioned below.

Setting μ(τ)=τ, w(τ)=1 and β=k=1, then we get  a+Iwμ1(x)=x-a and

Ra+f(x)=∫axf(t)dt, x>a.

Corollary 3.1. ( Reverse Minkowski type inequality via Riemann integral operator.) Let f,g>0,1≤q≤p<+∞,α>0 and

0<c<m≤αf(s)g(s)≤M,  for all s∈[a,x],

then

(3.9)M+αα(M−c)∫ax (αf(t)−cg(t))q dt1q≤∫axfq(t)dt1q +  x−ap−q q ∫axgp(t)dt1p≤m+αα(m−c) x−ap−q q  ∫ax (αf(t)−cg(t)) p dt1p.

The inequality (3.9) is a new result via Riemann operator on [a,x] with two parameters 0<q≤p and for q=p we get [1, Theorem 1.2].

Setting w(τ)=1 and μ(τ)=τ we get  a+Iwμ1(x)=1Γk(β+k)(x-a)βk and

RLa+kf(x)=1kΓk(β)∫ax (x−t) β k −1 f(t)dt, x>a.

Corollary 3.2. (Reverse Minkowski type inequality via k-Riemann-Liouville operator.) Under the assumptions of the Corollary 3.1 , we have

(3.10)M+αα(M−c)RL a+k (αf(x)−cg(x))q1q≤RL a+k fq(x)1q+ 1Γk(β+k) (x−a)βk p−qq  RL a+k gp(x)1p≤m+αα(m−c) 1Γk(β+k) (x−a)βk p−qq  RL a+k (αf(x)−cg(x))p1p,

The inequality (3.10) is a new result via the k-Riemann-Liouville operator on [a,x] with two parameters 0<p≤q, if we take k=1 we get a new Riemann-Liouville result.

Setting w(τ)=τ and μ(τ)=lnτ, we deduce  a+Iwμ1(x)=1Γk(β+k)lnxaβk and

Ha+kf(x)=1kΓk(β)∫axlnxtβk−1f(t)tdt, x>a>1.

Corollary 3.3. (Reverse Minkowski type inequality via k-Hadamard operator.) Under the assumptions of the Corollary 3.1 , we have

(3.11)M+αα(M−c) H a+k (αf(x)−cg(x))q1q≤ H a+k fq(x)1q+ 1 Γk(β+k) lnxa βk p−qq H a+k gp(x)1p≤m+αα(m−c) 1 Γk(β+k) lnxa βk p−qq H a+k (αf(x)−cg(x))p1p.

Inequality (3.11) is a new result via the k-Hadamard operator on [a,x] with two parameters 0<p≤q. If we put k=1 we get a new result with the Hadamard operator.

Setting w(τ)=τ and μ(τ)=τρ+1ρ+1, we get  a+Iwμ1(x)=1Γk(β+k)xρ+1-aρ+1ρ+1βk and

Ka+kf(x)=(ρ+1)1−βkkΓk(β)∫ax(xρ+1−tρ+1)βk−1tρf(t)dt, x>a.

Corollary 3.4. (Reverse Minkowski type inequality via k-Katugompola operator.) Under the assumptions of the Corollary 3.1 we have for all ρ>-1

(3.12)M+αα(M−c)K a+k (αf(x)−cg(x))q1q≤K a+k fq(x)1q+ 1Γk(β+k) xρ+1 −aρ+1 ρ+1 βk p−qq  K a+k gp(x)1p≤m+αα(m−c) 1Γk(β+k) xρ+1 −aρ+1 ρ+1 βk p−qq  K a+k (αf(x)−cg(x))p1p.

Inequality (3.12) is a new result via the k-Katugompola operator on [a,x] with two parameters 0<p≤q. If we put k=1 we get a new result with the Katugompola operator. Setting w(τ)=τ and μ(τ)=(τ-a)θθ, we have

 a+ℑwμ1(x)=1Γk(β+k) (x−a)θ θ βkandC1kf(x)=θ1−βkkΓk(β)∫ax (x−a)θ− (t−a)θ β k−1 f(t)(t−a)1−θdt, x>a.

Corollary 3.5. (Reverse Minkowski type inequality via fractional k-conformal integral operator.) Under the assumptions of the Corollary 3.1 , we get

(3.13)M+αα(M−c) C1k (αf(x)−cg(x))q1q≤ C1k fq(x)1q+ 1 Γk(β+k) (x−a)θ θ βk p−qq C1k gp(x)1p≤m+αα(m−c) 1 Γk(β+k) (x−a)θ θ βk p−qq C1k (αf(x)−cg(x))p1p.

Inequality (3.13) is a new result via the k-conformal operator on [a,x] with two parameters 0<p≤q. If we put k=1 we get a result with the conformal operator.