Extensions of Bullen-Type Inequalities Using Second Derivative Methods

Extensions of Bullen-Type Inequalities Using Second Derivative Methods Mehmet Zeki Sarıkaya1 1Department of Mathematics, Düzce University, Düzce, 81620, Türkiye Corresponding author: sarikay[email protected] ORCID IDs: First Author: 0000-0002-6165-9242 DOI : 10.5281/zenodo.18024313 Abstract In the present paper, our aim is to modify Dragomir’s method in a slightly different manner and, using a new approach, to establish novel Hadamard-type inequalities for two functions g and h , where the ratio of their second derivatives is bounded by the same lower bound m and upper bound M . Subsequently, we examine special cases of these inequalities through examples and, in certain instances, recover the results previously obtained by Dragomir et al. [4,5]. Keywords: Hermite-Hadamard inequaly, Bullen inequality, convex function 1 Introduction Integral inequalities of Hadamard and Bullen type play a fundamental role in approximation theory, numerical analysis, and convex analysis. Classical Hadamard inequalities provide bounds for the difference between the integral of a convex function and its midpoint or endpoint evaluations. Bullen-type inequalities generalize these results by comparing weighted sums of function values with the integral, often using second derivative information to establish sharp bounds. The Hermite-Hadamard inequality is stated as follows: Let f : I⊆R→R be a convex function on the interval I, and let a, b ∈Iwith a < b. Then, fa+b 2≤1 b−aZb a f(t)dt ≤f(a)+f(b) 2.(1) Inequality (1) can be further refined. For instance, the Bullen inequality provides a tighter upper bound as follows: fa+b 2≤1 b−aZb a f(t)dt ≤1 2f(a)+f(b) 2+fa+b 2.(2) Moreover, many authors have investigated new upper and lower bounds for the differences appearing in the Hermite-Hadamard inequality, typically expressed as  fa+b 2−1 b−aZb a f(x)dx ≤ · · · .(3) M26-1 KOSC-2025 Proceedings Inequalities of the form (3) are commonly referred to as midpoint inequalities. The exact representation of the difference can be expressed using the first derivative as fa+b 2−1 b−aZb a f(x)dx =1 b−a"Za+b 2 a (t−a)f′(t)dt +Zb a+b 2 (b−t)f′(t)dt#,(4) which was presented by Kırmacı in [7]. On the other hand, inequalities of the form  1 b−aZb a f(x)dx −f(a)+f(b) 2 ≤ · · · (5) are known as trapezoid inequalities. The corresponding exact expression can be written as 1 b−aZb a f(x)dx −f(a)+f(b) 2=1 b−aZb at−a+b 2f′(t)dt, (6) as given by Dragomir and Agarwal in [ 3 ]. Using the identities (4) and (6), many authors have derived various new midpoint and trapezoid type inequalities for different classes of convex functions; for more details, see references [ 3 ], [ 6 ], [ 7 ], [ 9 ]. These inequalities provide bounds for the integral mean of a function in terms of the function values at specific points, such as the midpoint or endpoints of an interval. Later, in [ 4 ] and [ 5 ], Dragomir and collaborators replaced the convexity assumption on f with a condition on the boundedness of its second derivative. Under this framework, they provided explicit lower and upper bounds for the differences in the midpoint (3) and (5) inequalities, establishing precise estimates for these cases.: Theorem 1.1. [ 4 , 5 ] Let f : [a, b]→R be a twice differentiable mapping such that there exists real constants m and M so that m≤f′′ ( t ) ≤M for t∈ ( a, b ). Then, the following inequalities hold: m(b−a)2 24 ≤1 b−a b Z a f(x)dx −fa+b 2≤M(b−a)2 24 .(7) and m(b−a)2 12 ≤f(a)+f(b) 2−1 b−a b Z a f(x)dx ≤M(b−a)2 12 .(8) In recent years, special versions of Hadamard-type inequalities have been developed for parametric classes of functions, and the influence of convexity conditions on the validity of these inequalities has been studied in detail. In particular, functions defined as various linear combinations of polynomials and power functions frequently arise in both theoretical and applied analyses. The sign and properties of their second derivatives serve as a primary tool for investigating convexity. For further details, see [ 1 ], [ 2 ], [ 8 ]–[ 10 ]. Recently, Sarikaya [ 9 ], building upon the approach of Dragomir et al. [ 4 , 5 ], utilized the Bullen inequality instead of the Hermite-Hadamard inequality to obtain more precise midpoint and trapezoid inequalities, and to derive several significant results concerning new approaches to these inequalities for two functions g and h , where the ratio of their second derivatives is bounded. In the present paper, our aim is to modify Dragomir’s method in a slightly different manner and, using a new approach, to establish novel inequalities with the same lower bound m and upper bound M . Subsequently, we M26-2 2nd Kocaeli Science Congress, November 19-21, 2025 examine special cases of these inequalities through examples and, in certain instances, recover the results previously obtained by Dragomir. 2 Main Theorem Theorem 2.1. Let g, h : [ a, b ] →R be twice continuously differentiable functions on [ a, b ], and assume that h′′(t)>0for all t∈[a, b]. Suppose there exist real constants mand Msuch that m≤g′′(t) h′′(t)≤M, ∀t∈[a, b]. Then the following two-sided bounds for the average of ghold. ga+b 2−m"ha+b 2−1 b−aZb a h(t)dt# ≤1 b−aZb a g(t)dt (9) ≤g(a)+g(b) 2−m"h(a)+h(b) 2−1 b−aZb a h(t)dt#. and g(a)+g(b) 2−M"h(a)+h(b) 2−1 b−aZb a h(t)dt# ≤1 b−aZb a g(t)dt (10) ≤ga+b 2−M"ha+b 2−1 b−aZb a h(t)dt#. Proof. Let g, h : [ a, b ] →R be twice continuously differentiable functions such that h′′ ( t ) > 0on [a, b]. Assume that there exist real constants mand Msatisfying m≤g′′(t) h′′(t)≤M, ∀t∈[a, b]. From m≤g′′(t)/h′′(t), we have g′′(t)−m h′′(t)≥0, t ∈[a, b]. Integrating this inequality over the symmetric interval [ t, a + b−t ](where a≤t≤ ( a + b ) / 2) yields Za+b−t tg′′(u)−m h′′(u)du ≥0, that is, g′(a+b−t)−g′(t)≥mh′(a+b−t)−h′(t). 2nd Kocaeli Science Congress, November 19-21, 2025 M26-3 KOSC-2025 Proceedings Integrating again with respect to tover [x, (a+b)/2], for x∈[a, (a+b)/2], gives Z(a+b)/2 xg′(a+b−t)−g′(t)dt ≥mZ(a+b)/2 xh′(a+b−t)−h′(t)dt. Evaluating these integrals yields g(a+b−x)+g(x)−2ga+b 2≥mh(a+b−x)+h(x)−2ha+b 2. Now integrate both sides over x∈[a, (a+b)/2]. By the symmetry property Z(a+b)/2 a [g(x)+g(a+b−x)] dx =Zb a g(x)dx, we obtain Zb a g(x)dx −(b−a)ga+b 2≥m"Zb a h(x)dx −(b−a)ha+b 2#. Dividing by (b−a)gives ga+b 2−m"ha+b 2−1 b−aZb a h(t)dt#≤1 b−aZb a g(t)dt. (11) Next, define φ ( t ) = g ( t ) −m h ( t ). Since φ′′ ( t ) = g′′ ( t ) −m h′′ ( t ) ≥ 0, the function φ is convex on [a, b]. By the classical Hermite-Hadamard inequality for convex functions, φa+b 2≤1 b−aZb a φ(t)dt ≤φ(a)+φ(b) 2. Substituting φ(t) = g(t)−m h(t)gives 1 b−aZb a g(t)dt ≤g(a)+g(b) 2−m"h(a)+h(b) 2−1 b−aZb a h(t)dt#.(12) Combining (11) and (12) yields the inequalities (9). Define the auxiliary function Ψ(t)=M h(t)−g(t), t ∈[a, b]. Then Ψ′′(t)=M h′′(t)−g′′(t)≥0, so Ψis convex on [a, b]. Applying the same symmetric integration argument as above gives Zb a Ψ(x)dx −(b−a) Ψa+b 2≥0, or equivalently, 1 b−aZb a Ψ(x)dx ≥Ψa+b 2. M26-4 2nd Kocaeli Science Congress, November 19-21, 2025 Substituting Ψ(t) = M h(t)−g(t)and rearranging yields 1 b−aZb a g(t)dt ≤ga+b 2−M"ha+b 2−1 b−aZb a h(t)dt#. Finally, applying the Hermite-Hadamard inequality to Ψ(t)gives Ψa+b 2≤1 b−aZb a Ψ(t)dt ≤Ψ(a) + Ψ(b) 2, which, after substituting Ψ(t)=M h(t)−g(t), leads to g(a)+g(b) 2−M"h(a)+h(b) 2−1 b−aZb a h(t)dt#≤1 b−aZb a g(t)dt. Together these give the inequalities (10), completing the proof. Corollary 2.1. Let f : [ a, b ] →R be twice continuously differentiable and assume there exist real constants m and M with m≤f′′ ( x ) ≤M for all x∈ [ a, b ]. Then, we have the following Bullen inequalities (2m−M)(b−a)2 48 ≤1 2f(a)+f(b) 2+fa+b 2−1 b−aZb a f(x)dx ≤(2M−m)(b−a)2 48 . (13) Proof. Apply Theorem 2.1 with g=fand with the auxiliary function h1(x) := 1 2x−a+b 22 , which satisfies h′′ 1(x)=1>0on [a, b]. Then f′′(x) h′′ 1(x)=f′′(x), so that the hypothesis of Theorem 2.1 is satisfied with the same constants m and M . By applying the Theorem 2.1, we obtain fa+b 2−m"h1a+b 2−1 b−aZb a h1(t)dt#≤1 b−aZb a f(t)dt. A simple computation gives h1a+b 2= 0,1 b−aZb a h1(t)dt =(b−a)2 24 . Hence, the lower bound becomes fa+b 2+m(b−a)2 24 ≤1 b−aZb a f(t)dt. (14) Similarly, the upper bound follows by applying the Hermite-Hadamard inequality to the concave 2nd Kocaeli Science Congress, November 19-21, 2025 M26-5 KOSC-2025 Proceedings function φ(x):=f(x)−Mh1(x), which yields 1 b−aZb a φ(t)dt ≤φa+b 2 and so 1 b−aZb a f(t)dt ≤fa+b 2+M(b−a)2 24 .(15) By combining inequalities (14) and (15), we obtain the midpoint-type inequality (7). To derive the trapezoidal estimate, we choose the auxiliary function h2(x) := 1 2(a−x)(b−x), x ∈[a, b]. We compute its second derivative: h′′ 2(x)=1>0. Then we consider f′′(x) h′′ 2(x)=f′′(x)∈[m, M], so the hypotheses of Theorem 2.1 are satisfied with constants m and M Now, let us compute the needed values; h2(a)=h2(b)=0, h2a+b 2=−(b−a)2 8,1 b−aZb a h2(x)dx =−(b−a)2 12 . By applying the Hermite-Hadamard inequality to the convex function φ(t) := f(t)−mh2(t) which yields 1 b−aZb a φ(t)dt ≤φ(a)+φ(b) 2 and thus 1 b−aZb a f(t)dt ≤f(a)+f(b) 2+m1 b−aZb a h2(t)dt =f(a)+f(b) 2−m(b−a)2 12 , f(a)+f(b) 2−1 b−aZb a f(x)dx ≥m(b−a)2 12 .(16) Similarly, the upper bound yields to the convex function φ ( t ) := Mh2 ( t ) −f ( t )and a simple computation f(a)+f(b) 2−1 b−aZb a f(x)dx ≤M(b−a)2 12 .(17) Therefore, by combining inequalities (16) and (17), we obtain the trapezoidal-type inequality, which corresponds to (8). Let A := f(a)+f(b) 2−1 b−aRb af(t)dt and B := 1 b−aRb af(t)dt − fa+b 2 . Then the target expression equals 1 2 ( A−B )in (13). Using the displayed bounds for M26-6 2nd Kocaeli Science Congress, November 19-21, 2025 Aand Bsuch as m(b−a)2 12 ≤A≤M(b−a)2 12 and m(b−a)2 24 ≤B≤M(b−a)2 24 we obtain m 12 −M 24(b−a)2≤A−B≤M 12 −m 24(b−a)2, i.e. (2m−M)(b−a)2 24 ≤A−B≤(2M−m)(b−a)2 24 . Dividing by 2yields the asserted inequality (2m−M)(b−a)2 48 ≤1 2(A−B)≤(2M−m)(b−a)2 48 , This completes the proof of Corollary 2.1. Corollary 2.2. Let g : [ a, b ] →R be twice continuously differentiable on [ a, b ]. Assume there exist constants mand Msuch that m≤g′′(t)≤M, ∀t∈[a, b]. Then the inequalities of Theorem 2.1 reduce (for h ( x ) = x2 2 ) to the following Hermite-Hadamard type bounds for the mean value of g. ga+b 2+m 24(b−a)2≤1 b−aZb a g(t)dt ≤g(a)+g(b) 2−m 12(b−a)2. and g(a)+g(b) 2−M 12(b−a)2≤1 b−aZb a g(t)dt ≤ga+b 2+M 24(b−a)2. In particular, ga+b 2+m 24(b−a)2≤1 b−aZb a g(t)dt ≤ga+b 2+M 24(b−a)2. Proof. Take h(x) = x2 2. Then h′′ (x)=1,h(a) = a2/2,h(b)=b2/2, and so ha+b 2=(a+b)2 8,1 b−aZb a h(t)dt =a2+ab +b2 6. From Theorem 2.1, we get ga+b 2−m"ha+b 2−1 b−aZb a h(t)dt#≤1 b−aZb a g(t)dt. By computation, it follows that ha+b 2−1 b−aZb a h(t)dt =(a+b)2 8−a2+ab +b2 6=−(b−a)2 24 . 2nd Kocaeli Science Congress, November 19-21, 2025 M26-7 KOSC-2025 Proceedings Hence ga+b 2+m 24(b−a)2≤1 b−aZb a g(t)dt. From Theorem 2.1, we have 1 b−aZb a g(t)dt ≤g(a)+g(b) 2−m"h(a)+h(b) 2−1 b−aZb a h(t)dt#. By computation, it follows that h(a)+h(b) 2−1 b−aZb a h(t)dt =a2+b2 4−a2+ab +b2 6=(a−b)2 12 =(b−a)2 12 , so 1 b−aZb a g(t)dt ≤g(a)+g(b) 2−m 12(b−a)2. The M-side is obtained by the same substitutions with Min place of m: g(a)+g(b) 2−M 12(b−a)2≤1 b−aZb a g(t)dt ≤ga+b 2+M 24(b−a)2. Combining these yields the displayed inequalities. Theorem 2.2. Let a, b > 0with a<b , let p > 2, and let g : [ a, b ] →R be twice continuously differentiable. Suppose there exist constants mand Msuch that m tp−2≤g′′ (t)≤M tp−2,∀t∈[a, b]. Then the following Hadamard-type inequalities hold: ga+b 2−m p(p−1) "a+b 2p −bp+1 −ap+1 (p+ 1)(b−a)# ≤1 b−aZb a g(t)dt ≤g(a)+g(b) 2−m 2p(p−1) "ap+bp−2(bp+1 −ap+1) (p+ 1)(b−a)#. and g(a)+g(b) 2−M 2p(p−1) "ap+bp−2(bp+1 −ap+1) (p+ 1)(b−a)# ≤1 b−aZb a g(t)dt ≤ga+b 2−M p(p−1) "a+b 2p −bp+1 −ap+1 (p+ 1)(b−a)#. Proof. Set h ( t ) := tp/ ( p ( p− 1)), so that h′′ ( t ) = tp−2> 0. Then the assumption m tp−2≤ g′′ (t)≤M tp−2is equivalent to g′′(t)−m tp−2≥0, M tp−2−g′′(t)≥0. M26-8 2nd Kocaeli Science Congress, November 19-21, 2025 Applying the symmetric integration method from Theorem 2.1, for the m-block we obtain ga+b 2−m"ha+b 2−1 b−aZb a h(t)dt# ≤1 b−aZb a g(t)dt ≤g(a)+g(b) 2−m"h(a)+h(b) 2−1 b−aZb a h(t)dt#. Similarly, for the M, applying the same integration procedure to Mtp−2−ggives g(a)+g(b) 2−M"h(a)+h(b) 2−1 b−aZb a h(t)dt# ≤1 b−aZb a g(t)dt ≤ga+b 2−M"ha+b 2−1 b−aZb a h(t)dt#. Finally, substituting h(a) = ap p(p−1), h(b) = bp p(p−1), h a+b 2=(a+b)p 2pp(p−1),Zb a h(t)dt =bp+1 −ap+1 (p+ 1)p(p−1) yields the explicit inequalities in the statement. Theorem 2.3. Let a, b > 0with a<b , let p > 2, and let g : [ a, b ] →R be twice continuously differentiable. Suppose there exist constants mand Msuch that ma+b 2−tp−2 ≤g′′(t)≤Ma+b 2−tp−2 , t ∈[a, b]. Then the following Hadamard-type inequalities hold: ga+b 2+m(b−a)p 2p(p+ 1)p(p−1) ≤1 b−aZb a g(t)dt ≤g(a)+g(b) 2−m(b−a)p 2pp(p−1) +m(b−a)p 2p(p+ 1)p(p−1). and g(a)+g(b) 2−M(b−a)p 2pp(p−1) +M(b−a)p 2p(p+ 1)p(p−1) ≤1 b−aZb a g(t)dt ≤ga+b 2+M(b−a)p 2p(p+ 1)p(p−1). Proof. Let a, b > 0with a < b, let p > 2, and define h(t) := a+b 2−tp p(p−1) , t ∈[a, b]. 2nd Kocaeli Science Congress, November 19-21, 2025 M26-9