Definition 1 (see [13], [14].)Consider f ∈ L[a, b], then the RHS and the LHS Riemann-Liouville fractional integral (RL) of order α > 0 with b > a > 0 are defined by

respectively, where Γ(α) is the Gamma function defined as

It is to be noted that $$

Lemma 2 (see [19].)Let p be any nonzero real number and α be any positive constant. Further consider an integrable function w : A⟶ℝ, where A = [a, b] ⊂ (0, ∞) which is p-symmetric w.r.t. $$; then, we have the following:

• (i)

  If p > 0,

with g(x) = x^1/p, x ∈ [a^p, b^p]

• (ii)

  If p < 0,

with g(x) = x^1/p, x ∈ [b^p, a^p]

Theorem 3. Let the strongly generalized p-convex function f : I⟶ℝ with magnitude μ > 0 and η (·) be bounded above in f(I) × f(I) y f ∈ L[a, b]. Then, if p is any positive real number, we have

Proof. We begin the proof by inserting $$ and $$

Take $$ and $$, then (8) yields

Multiplying (9) by t^α−1 and then integrating w.r.t. t over the interval [0, 1/2],

which is the left side of Theorem 3

Now, to obtain the left-hand side of Theorem 3, we have for $$,

and for $$

Combining (12) and (13), we have

Multiplying (14) by 2t^α−1 and then integrating w.r.t. t over the interval [0, 1/2], we have

Together (11) and (16) give the required result.

Remark 4.

• (i)

  Fixing p = 1 in Theorem 3 gives Hermite-Hadamard inequality in the sense of the strongly generalized convexity

• (ii)

  Fixing p = 1 and μ = 0 in Theorem 3, we obtain [20] (Theorem 2.1)

• (iii)

  Fixing η(x, y) = x − y and μ = 0 in Theorem 3 yields [21] (Theorem 2.1)

• (iv)

  Applying both (ii) and (iii) on Theorem 3, we obtain classical fractional version of H-H inequality

Definition 5 (see [22].)Let p be any nonzero real number; then, the function w : [a, b]⟶ℝ is p-symmetric w.r.t. $$ if $$ for all x ∈ [a, b].

Theorem 6 (inequality of Fejér type). Suppose that f is a function as in Theorem 3 and an integrable, nonnegative function w : [a, b]⟶R is symmetric w.r.t. $$, then

Proof. Setting t = 1/2 in (2),

Substitute $$ and $$ in (18),

According to the given conditions of w, we have

∀x, y ∈ [a, b]. Multiplying (19) by $$ and then integrating w.r.t. t over the interval [0, 1],

Let g(x) = x^1/p, then (23) becomes

Now, take x = (ta^p + (1 − t)b^p)∀t ∈ [0, 1], then by Def of f,

Multiply on both sides of (26) by $$ and then integrate w.r.t. t over the interval [0, 1],

Take g(x) = x^1/p in (28), then we have

Similarly, we have

from definition of f by fixing x = tb^p + (1 − t)a^p.

Combining (29) and (30), we obtain

Combining (32) and (25) completes the theorem (17).

Lemma 7. Consider a differentiable function f : I ⊂ (0, ∞)⟶R on I^o, with f^′ ∈ L[a, b], where a, b ∈I and a < b. If w : [a, b]⟶R is integrable, then

holds with g(x) = x^1/p.

Proof. Let p > 0, and x∈[a^p, b^p], then for generalized strongly p-convex function, we have

where

By integration by parts, we have

By combining (34), (37), and (38), we have (33). This completes the proof.

Remark 8. Setting μ = 0 and η = x − y in Lemma 7 gives us [21] (Lemma 2.1).

Theorem 9. Let the function f be as in Theorem 3.1. If |f^′| is a strongly generalized p-convex function on [a, b] for positive p and α, then

where

Proof. Theorem (3) gives

Setting t = ua^p + (1 − u)b^p, dt = (a^p − b^p)du, we have

Since |f^′| is a strongly generalized p-convex function on [a, b], we have

After combining (48) and (43), we have

Remark 10. If one takes η = x − y and μ = 0, then we get [21] (Theorem 2.2).

Theorem 11. Let the function f be as in Theorem 3.1. If |f^′| is as in Theorem 9, then

where

Proof. Let p > 0:

Setting t = ua^p + (1 − u)b^p, dt = (a^p − b^p)du, we have

Using the inequality of power mean the definition of $$,

This completes the proof.

Remark 12.

• (i)

  Setting p = 1 and μ = 0, η = xy in Theorem 11 gives H-H type inequality for convex functions

• (ii)

  Setting p = 1 and α = 1, μ = 0, η = xy in Theorem 11 gives H-H inequality for convex functions