Page 163 - IJOCTA-15-1

Improvements of Hermite-Hadamard-Mercer inequality using k-fractional integral
α
ν : [δ 1 , δ 2 ] → R be integrable, nonnegative and Multiplying (8) with α t κ −1 and integrating from
κ
symmetric to δ 1 +δ 2 . Then the fractional integrals 0 to 1, we get
2
inequalities
αu + κv
ω(δ 1 + δ 2 − )

δ 1 + δ 2 γ γ α + κ
ω J + ν(δ 2 ) + J − ν(δ 1 )
2 δ 1 δ 2 ≤ ω(δ 1 ) + ω(δ 2 )
γ
γ
≤ J + (ων) (δ 2 ) + J − (ων) (δ 1 ) 2 Z 1
δ δ α α −1 κt κt
1 2 − t κ ω (1− )u+ v dt
ω(δ 1 ) + ω(δ 2 ) γ γ κ(α + κ) 0 α + κ α + κ
≤ J + ν(δ 2 ) + J − ν(δ 1 ) Z
2 δ 1 δ 2 α 1 α −1 αt αt
− t κ ω u + (1 − )v dt
α + κ 0 α + κ α + κ
hold for all γ > 0. αu + κv
⇒ ω(δ 1 + δ 2 − ) ≤ ω(δ 1 ) + ω(δ 2 )
Adil et. al in, 42 gave H-H-Mercer type inequalities α + κ α
for strongly convex functions. They also estab- α 2 Z αu+κv α + κ λ − u κ −1
α+κ
−
lished H-H-mercer type inequalities for differen- κ(α + κ) u κ v − u
tiable functions. In this article, we present H-H- α + κ dλ
Mercer type inequality, involving both right and × ω(λ) κ v − u
left k-fractional integral operators. Z αu+κv α −1
α α+κ α + κ v − η κ
−
2. Main result α + κ v α v − u

α + κ dη
Theorem 3. Let ω : [δ 1 , δ 2 ] → R be convex func- × ω(η) −
tion, u, v ∈ [δ 1 , δ 2 ] such that u < v. Then the α v − u
inequalities
αu + κv
⇒ ω(δ 1 + δ 2 − )
α + κ

αu + κv
ω δ 1 + δ 2 − ≤ ω(δ 1 ) + ω(δ 2 )
α + κ
α
2
α −1 α (α + κ) κ −1 Z αu+κv α
α+κ
(α + κ) κ Γ κ (α) − (λ − u) κ −1 ω(λ)dλ
α
α
≤ ω(δ 1 ) + ω(δ 2 ) − α κ κ +1 (v − u) κ
(v − u) κ u
α Z
2 (α + κ) κ −1 v α
α α,κ κ α,κ − −1
× α J αu+κv − ω(u) + α J αu+κv + ω(v) α −1 α (v − η) κ ω(η)dη
κ κ ( α+κ ) α κ −1 ( α+κ ) α κ (v − u) κ αu+κv
α+κ

αu + κv
≤ ω(δ 1 ) + ω(δ 2 ) − ω (6) αu + κv
α + κ ⇒ ω(δ 1 + δ 2 − )
α + κ
2 α −1
hold, where α, κ > 0. α Γ κ (α)(α + κ) κ α,κ
≤ ω(δ 1 )+ω(δ 2 )− α α J αu+κv − ω(u)
κ κ (v − u) κ ( α+κ )
Proof. From Mercer’s inequality, we have
α −1
κΓ κ (α)(α + κ) κ α,κ
− α −1 α J αu+κv + ω(v)
αx + κy α κ (v − u) κ ( α+κ )
ω(δ 1 + δ 2 − )
α + κ
α κ αu + κv
≤ ω(δ 1 ) + ω(δ 2 ) − ω(x) − ω(y) (7) ⇒ ω(δ 1 + δ 2 − )
α + κ α + κ α + κ
α 2
Γ κ (α)(α + κ) κ −1 α α,κ
Substituting x = (1 − κt )u + ( κt )v and y = ≤ ω(δ 1 )+ω(δ 2 )− α α J αu+κv − ω(u)
α+κ α+κ (v − u) κ κ κ ( α+κ )
αt u + (1 − αt
α+κ α+κ )v, where t ∈ [0, 1] we have κ α,κ
+ α J αu+κv + ω(v) . (9)
α κ −1 ( α+κ )
αu + κv
ω(δ 1 + δ 2 − ) Now we prove the second part of inequality (6).
α + κ

α κt κt
≤ ω(δ 1 )+ω(δ 2 )− ω (1− )u+( )v αu + κv
α + κ α + κ α + κ ω =
α + κ

κ αt αt  κt κt αt αt 
− ω u + (1 − )v . (8) α α+κ v + 1 − α+κ u + κ 1 − α+κ v + α+κ u
α + κ α + κ α + κ ω  
α + κ
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