The Hermite-Hadamard inequality revisited: Some new proofs and applications
aa r X i v : . [ m a t h . G M ] O c t Noname manuscript No. (will be inserted by the editor)
The Hermite-Hadamard inequality revisited: Somenew proofs and applications
Ilham A. Aliev · Mehmet E. Tamar · Cagla Sekin
Received: date / Accepted: date
Abstract
New proofs of the classical Hermite-Hadamard inequality are pre-sented and several applications are given, including Hadamard-type inequali-ties for the functions, whose derivatives have inflection points or whose deriva-tives are convex. Morever, some estimates from below and above for the firstmoments of functions f : [ a, b ] → R about the center point c = ( a + b ) / f : [0 , ∞ ) → (0 , ∞ ) is established. Keywords
Convex functions · Hermite-Hadamard inequality · Jenseninequality · Fejer inequality.
Mathematics Subject Classification (2010) · · The famous Hermite-Hadamard inequality (HH) asserts that the integral meanvalue of a convex function f : [ a, b ] → R can be estimated above and beloveby its values at the points a, b and ( a + b ) /
2. More precisely, f (cid:18) a + b (cid:19) ≤ b − a Z ba f ( x ) dx ≤ f ( a ) + f ( b )2 . (HH) I. A. ALIEVDepartment of mathematics, Akdeniz University, Antalya, TURKEYE-mail: [email protected]. E. TAMARInstitute of science, Akdeniz University, Antalya, TURKEYC. SEKINInstitute of science, Akdeniz University, Antalya, TURKEY Ilham A. Aliev et al.
Equality holds only for functions of the form f ( x ) = cx + d . According tothe notations by Niculescu and Persson [20], the right and left parts of (HH)we denote by (RHH) and (LHH), respectively. Some authors also refer to the(RHH) as Hadamard’s inequality.(HH) has many generalizations, extensions, refinements and there is alarge number of papers and book’s chapters in this area; see, e.g. books byNiculescu and Persson [21]; Mitrinovic, Pecaric and Fink [19]; Dragomir andPearce [9] and papers [1,2,3,4,5,6,7,8,10,11,12,13,14,15,16,17,18,20,22,23,24,25,26,27], which are a small part of the relevant references.The plan of this article is as follows.In section 2 we give two new proofs of (HH), one of which is extremely short(so to speak, ”without pulling the pen out of paper”), and the other is basedon Riemann’s integral sums. As an application, we give an estimation frombelow and above of the integral of the convex function f : [0 , ∞ ) → (0 , ∞ ) viathe series ∞ P f ( k ) and ∞ P f (cid:0) k − (cid:1) . In section 3, we give some new inequalities arising as a combination of the(HH) with the Hardy inequality and some variant of H¨older’s inequality. Forexample, as a consequence we prove that, if f : [0 , ∞ ) → (0 , ∞ ) is convex and f ∈ L p (0 , ∞ ), ∀ p >
1, then lim p →∞ (cid:13)(cid:13) x R x f (cid:13)(cid:13) p k f k p = 1 . Section 4 is devoted to Hadamard’s type inequality, i.e. (RHH) for thefunctions whose first derivatives have an inflection point. As a particular case,we show that if f ′ is concave on (cid:2) a, a + b (cid:3) and convex on (cid:2) a + b , b (cid:3) , then f ( a ) + f ( b )2 − b − a Z ba f ( x ) dx ≤ b − a
12 ( f ′ ( b ) − f ′ ( a )) . In the last 5th section we prove various inequalities for functions havingconvex first or second order derivatives. As far as we know, in the literatureon this theme there are some inequalities for functions whose absolute values of the derivatives are convex, see, e.g. [2,6,24]. In our Theorem 7, 8, and 9 insection 5, the convexity condition is imposed on the derivatives themselves, butnot on their absolute values. One of the interesting particular results obtainedin this section is as follows.Given f : [ a, b ] → R , let f ′ be convex. Then Z b a + b f ( x ) dx − Z a + b a f ( x ) dx ≤ b − a f ( b ) − f ( a )) . Another result in this section is the estimation from above and below of thefirst moment of a function f : [ a, b ] → R about the center point c = ( a + b ) / M f = R ba (cid:0) x − a + b (cid:1) f ( x ) dx , when f ′ is convex. he Hermite-Hadamard inequality revisited: Some new proofs and applications 3 At first, we give an auxiliary inequality that is satisfied by convex functions.
Lemma 1 (cf. Lemma 1.3 in [18]) Let f be a convex function on [ a, b ] . Then f ( a ) + f ( b ) ≥ f ( a + b − x ) + f ( x ) , ( ∀ x ∈ [ a, b ] ). (1) Proof
Let x ∈ [ a, b ]. There exists a t ∈ [0 ,
1] such that x = ta + (1 − t ) b . Then a + b − x = (1 − t ) a + tb and therefore, f ( a + b − x ) = f ((1 − t ) a + tb ) ≤ (1 − t ) f ( a ) + tf ( b )= f ( a ) + f ( b ) − [ tf ( a ) + (1 − t ) f ( b )] ≤ f ( a ) + f ( b ) − f ( ta + (1 − t ) b )= f ( a ) + f ( b ) − f ( x ) . By making use of (1) we give here a short proof of the (HH), ”withoutpulling the pen out of paper”.
Proof of (HH)
Integrating the inequality f ( a ) + f ( b ) ≥ f ( a + b − x ) + f ( x )over [ a, b ] and using Z ba f ( a + b − x ) dx = Z ba f ( x ) dx we have( b − a )( f ( a ) + f ( b )) ≥ Z ba f ( x ) dx = Z ba ( f ( a + b − x ) + f ( x )) dx = 2 Z ba f ( a + b − x ) + f ( x )2 dx ≥ Z ba f (cid:18) a + b − x + x (cid:19) dx = 2( b − a ) f (cid:18) a + b (cid:19) . Remark 1
Although the (HH) has several proofs, as far as we know the firstsimple proof was given by Azbetia [3]; (see, also Niculescu and Persson [20],p. 664). Another simple proof and refinement was given by El Farissi [11].
Remark 2
The inequality (1) enables one also to give a short proof of ”Hadamardpart” of the Fejer inequality:If g ≥ a + b , i.e. g ( a + b − x ) = g ( x ), ( x ∈ [ a, b ]), we have Z ba f ( x ) g ( x ) dx = 12 "Z ba f ( x ) g ( x ) dx + Z ba f ( a + b − x ) g ( a + b − x ) dx = 12 Z ba ( f ( x ) + f ( a + b − x )) g ( x ) dx ( ) ≤ f ( a ) + f ( b )2 Z ba g ( x ) dx. Ilham A. Aliev et al.
The following theorem is the ”Riemann integral’s sums version” of (HH).
Theorem 1 If f : [ a, b ] → R is convex and x k = a + k b − an , ( k = 1 , , · · · , n )then for any n ∈ N we have f (cid:0) − n (cid:1) a + (cid:0) n (cid:1) b ! ≤ n n X k =1 f ( x k ) ≤ (cid:20) f ( a ) (cid:18) − n (cid:19) + f ( b ) (cid:18) n (cid:19)(cid:21) . (2) Proof
Let x k = a + k b − an , ( k = 1 , , · · · , n ). Then writing x k as x k = b − x k b − a a + x k − ab − a b and using f ( x k ) ≤ b − x k b − a f ( a ) + x k − ab − a f ( b ) , one has n X k =1 f ( x k ) ≤ f ( a ) b − a n X k =1 ( b − x k ) + f ( b ) b − a n X k =1 ( x k − a )= 12 [ f ( a )( n −
1) + f ( b )( n + 1)] , and therefore,1 n n X k =1 f ( x k ) ≤ (cid:20) f ( a ) (cid:18) − n (cid:19) + f ( b ) (cid:18) n (cid:19)(cid:21) . (3)On the other hand, the Jensen inequality yields1 n n X k =1 f ( x k ) ≥ f n n X k =1 x k ! = f (cid:0) − n (cid:1) a + (cid:0) n (cid:1) b ! . (4)By combining (3) and (4) we obtain (2). Corollary 1
After taking limit as n → ∞ in (2) and using the fact that theconvex function is continuous, we obtain (HH). The following two theorems are the simple consequences of (HH).
Theorem 2 (a refinement of (RHH))
Let f : [ a, b ] → R be convex. Then b − a Z ba f ( x ) dx ≤ b − a Z ba f ( x ) (cid:20) ln ( b − a ) ( b − x )( x − a ) − (cid:21) dx ≤ f ( a ) + f ( b )2 (5) he Hermite-Hadamard inequality revisited: Some new proofs and applications 5 Proof
For any x ∈ ( a, b ] one has f (cid:18) a + x (cid:19) ≤ x − a Z xa f ( t ) dt ≤ f ( a ) + f ( x )2 . Integrating over ( a, b ) we have Z ba f (cid:18) a + x (cid:19) dx ≤ Z ba x − a (cid:18)Z xa f ( t ) dt (cid:19) dx ≤ Z ba f ( a ) + f ( x )2 dx. (6)After simple calculations, (6) leads to2 Z a + b a f ( x ) dx ≤ Z ba f ( x ) ln b − ax − a dx ≤ " f ( a )( b − a ) + Z ba f ( x ) dx . (7)Similarly, integrating the inequality f (cid:18) x + b (cid:19) ≤ b − x Z ba f ( t ) dt ≤ f ( x ) + f ( b )2over ( a, b ) we get Z ba f (cid:18) x + b (cid:19) dx ≤ Z ba b − x Z ba f ( t ) dt ! dx ≤ Z ba f ( x ) + f ( b )2 dx which leads to2 Z b a + b f ( x ) dx ≤ Z ba f ( x ) ln b − ab − x dx ≤ " f ( b )( b − a ) + Z ba f ( x ) dx . (8)After summing up (7) and (8) we obtain (5). Theorem 3
Let f : [0 , ∞ ) → (0 , ∞ ) be a strictly convex function and ∞ P k =1 f ( x k ) < ∞ . Then ∞ X k =1 f (cid:18) k − (cid:19) < Z ba f ( x ) dx < f (0) + ∞ X k =1 f ( k ) . (9) Proof
Denote x = a and x k = a + ( k − b − an , ( k = 1 , , · · · , n ). Since f isstrictly convex, we have f (cid:18) x k − + x k (cid:19) < x k − x k − Z x k x k − f ( x ) dx < f ( x k − ) + f ( x k )2 , ( k = 1 , , · · · , n ) . Taking into account the formulas x k − x k − = b − an , x k − + x k a + (cid:18) k − (cid:19) b − an Ilham A. Aliev et al. and summing the inequalities above we obtain n X k =1 n f (cid:18) a + (cid:18) k − (cid:19) b − an (cid:19) < b − a Z ba f ( x ) dx< n " f ( a ) + f ( b )2 + n − X k =1 f (cid:18) a + k b − an (cid:19) Further, setting a = 0, b = n we have n X k =1 f (cid:18) k − (cid:19) < Z ba f ( x ) dx < f (0) + f ( n )2 + n − X k =1 f ( k ) . Taking limit as n → ∞ and using lim n →∞ f ( n ) = 0 we obtain the desired formula(9). Remark 3
Since f : [0 , ∞ ) → (0 , ∞ ) is convex and lim n →∞ f ( n ) is finite (actually,zero), then f is monotonically decreasing and therefore the comparison of theareas under graphics gives ∞ X k =1 f ( k ) < Z ∞ f ( x ) dx < f (0) + ∞ X k =1 f ( k ) . (10)It is clear that, the inequalities (9) are better than (10). Example 1 If f ( x ) = e − x then from (9) we have √ ee − < <
12 + 1 e − ⇔ √ e < e − <
12 ( e + 1) , whereas the formula (10) gives the rougher estimate 1 < e − < e . Let < p < ∞ and αp > . Let further, f : [0 , ∞ ) → (0 , ∞ ) beconvex and such that (cid:13)(cid:13) x − α f ( x ) (cid:13)(cid:13) p ≡ (cid:18)Z ∞ (cid:0) x − α f ( x ) (cid:1) p dx (cid:19) /p < ∞ . Then − α + p ≤ (cid:13)(cid:13) x − α R x f (cid:13)(cid:13) p k x − α f ( x ) k p ≤ α − /p . (11) he Hermite-Hadamard inequality revisited: Some new proofs and applications 7 Corollary 2 (a) If α = 1 , then p ≤ (cid:13)(cid:13) x R x f (cid:13)(cid:13) p k f k p ≤ − /p. (12) (b) Let, in addition, f ∈ L p (0 , ∞ ) , ( ∀ p > ). Then by taking the limit in(12) as p → ∞ one has lim p →∞ (cid:13)(cid:13) x R x f (cid:13)(cid:13) p k f k p = 1 . (13) Proof
We will use the classical weighted Hardy inequality, which asserts that (cid:18)Z ∞ (cid:12)(cid:12)(cid:12)(cid:12) x − α Z x f ( t ) dt (cid:12)(cid:12)(cid:12)(cid:12) p dx (cid:19) /p ≤ c (cid:18)Z ∞ (cid:12)(cid:12) x − α f ( x ) (cid:12)(cid:12) p dx (cid:19) /p , (14)where c = pαp − , 1 < p < ∞ , αp > . Now, by (LHH) we have f (cid:16) x (cid:17) < x Z x f ( t ) dt ⇒ x − α f (cid:16) x (cid:17) < x − α Z x f ( t ) dt, and therefore Z ∞ (cid:16) x − α f (cid:16) x (cid:17)(cid:17) p dx ≤ Z ∞ (cid:18) x − α Z x f ( t ) dt (cid:19) p dx ( ) ≤ (cid:18) pαp − (cid:19) p Z ∞ ( x − α f ( x )) p dx. (15)Since Z ∞ (cid:16) x − α f (cid:16) x (cid:17)(cid:17) p dx = 2 p (1 − α )+1 Z ∞ (cid:0) x − α f ( x ) (cid:1) p dx, we have from (15) the desired result (11) and its consequences (12) and (13). Remark 4
The left hand side of (11) shows that under the conditions of The-orem 4 the following reverse Hardy’s inequality is valid: (cid:13)(cid:13)(cid:13)(cid:13) x − α Z x f (cid:13)(cid:13)(cid:13)(cid:13) p ≥ − α + p (cid:13)(cid:13) x − α f ( x ) (cid:13)(cid:13) p . Example 2
Let k > f ( x ) = e − kx . Then (13) yieldslim p →∞ (cid:18)Z ∞ (cid:18) − e − kx x (cid:19) p dx (cid:19) /p = k. Ilham A. Aliev et al.
In the next theorem we will make use of a combination of Hadamard’sinequality 1 b − a Z ba f ( x ) dx ≤ f ( a ) + f ( b )2and the inequality Z ba n Y k =1 u k ( x ) ! dx ! n ≤ n Y k =1 Z ba u nk ( x ) dx ! , (16)where u ≥ , · · · , u n ≥ . Recall that the inequality (16) is a consequence of H¨older’s inequality andcan be proved by induction.We need also the following
Lemma 2 If u : [ a, b ] → (0 , ∞ ) is convex, then u n is convex as well for any n ∈ N . A simple proof follows by induction. Namely, if the functions u > u n are convex for some n ≥
2, i.e. u ( αx + βy ) ≤ αu ( x ) + βu ( y )and u n ( αx + βy ) ≤ αu n ( x ) + βu n ( y ), ( α + β = 1) , then by multiplying these inequalities we have u n +1 ( αx + βy ) ≤ ( αu ( x ) + βu ( y )) ( αu n ( x ) + βu n ( y )) ≤ αu n +1 ( x ) + βu n +1 ( y ) , where the last estimate is equivalent to the following obvious inequality:( u ( x ) − u ( y )) ( u n ( x ) − u n ( y )) ≥ . Remark 5
The convexity of the functions u ≥ , u ≥ , · · · , u n ≥ u u · · · u n . Indeed, for example,although the functions u ( x ) = x , u ( x ) = x , · · · , u n − ( x ) = x and u n ( x ) =(2 − x ) n − , ( n ≥
2) are convex on [0 , u ( x ) = x n − (2 − x ) n − is not convex because of f ′′ (1) = 4(2 n − − n ) < . Theorem 5
For given n ≥ , let the functions u ≥ , u ≥ , · · · , u n ≥ beconvex on [ a, b ] . Then b − a Z ba n Y k =1 u k ( x ) ! dx ≤ n Y k =1 ( u nk ( a ) + u nk ( b )) n . (17) he Hermite-Hadamard inequality revisited: Some new proofs and applications 9 Proof
Since u k , ( k = 1 , , · · · , n ) is convex on [ a, b ], then u nk is also convex byLemma 2. Then Hadamard’s inequality yields1 b − a Z ba u nk ( x ) dx ≤
12 [ u nk ( a ) + u nk ( b )] , ( k = 1 , , · · · , n ) . By multiplying these inequalities we have1( b − a ) n n Y k =1 Z ba u nk ( x ) dx ! ≤ n n Y k =1 ( u nk ( a ) + u nk ( b )) . (18)Here, by making use of the inequality (16), we get1( b − a ) n Z ba n Y k =1 u k ( x ) ! dx ! n ≤ n n Y k =1 ( u nk ( a ) + u nk ( b )) , from which the inequality (17) follows. Remark 6
For n = 2, the inequality (17) was proved by Amrahov [1]. Anothergeneralization of Amrahov’s result for the product of two functions was notedby D. A. Ion [16]:If u ≥ , v ≥ p + q = 1, (1 < p, q < ∞ ), then1 b − a Z ba u ( t ) v ( t ) dt ≤
12 ( u p ( a ) + u p ( b )) /p ( u q ( a ) + u q ( b )) /q . It should also be mentioned that, in the same paper [16] Ion gives some gener-alization of Amrahov’s result for the product of two functions in Orlicz spaces.
Given c ∈ [ a, b ] and f : [ a, b ] → R , let its derivative f ′ be concaveon [ a, c ] and convex on [ c, b ] . Then (cid:20) c − ab − a f ( a ) + b − cb − a f ( b ) (cid:21) − b − a Z ba f ( x ) dx ≤ (cid:20) ( b − c ) b − a f ′ ( b ) − ( c − a ) b − a f ′ ( a ) + (cid:18) a + b − c (cid:19) f ′ ( c ) (cid:21) . (19) Corollary 3
In case of c = a + b we have f ( a ) + f ( b )2 − b − a Z ba f ( x ) dx ≤ b − a
12 ( f ′ ( b ) − f ′ ( a )) (20) Proof
Integration by parts yields c − ab − a f ( a ) + b − cb − a f ( b ) − b − a Z ba f ( x ) dx = 1 b − a Z ba ( x − c ) f ′ ( x ) dx = 1 b − a Z ca ( x − c ) f ′ ( x ) dx + 1 b − a Z bc ( x − c ) f ′ ( x ) dx ≡ A + B. (21)By changing variables as x = (1 − λ ) a + λc , (0 < λ <
1) in A and x =(1 − λ ) c + λb in B and applying Jensen’s inequality, we have A ≡ b − a Z ca ( x − c ) f ′ ( x ) dx = ( a − c ) b − a Z ( λ − f ′ ((1 − λ ) a + λc ) dλ ≤ ( a − c ) b − a Z ( λ −
1) [(1 − λ ) f ′ ( a ) + λf ′ ( c )] dλ = − ( a − c ) b − a ) [2 f ′ ( a ) + f ′ ( c )] ; (22) B ≡ b − a Z bc ( x − c ) f ′ ( x ) dx = ( b − c ) ( b − a ) Z λf ′ ((1 − λ ) c + λb ) dλ ≤ ( b − c ) ( b − a ) Z (cid:0) λ (1 − λ ) f ′ ( c ) + λ f ′ ( b ) (cid:1) dλ = ( b − c ) b − a ) [ f ′ ( c ) + 2 f ′ ( b )] . (23)It follows from (22) and (23) that A + B ≤ f ′ ( b ) ( b − c ) b − a − f ′ ( a ) ( a − c ) b − a + 16 f ′ ( c )( a + b − c ) , which completes the proof. Remark 7
A simple calculation shows that the equality in (19) holds for thefunctions f ( x ) = k ( x − c ) + m , ( k, m ∈ R ). Remark 8
In the ”critic” cases c = a or c = b , i.e. in the cases when f ′ isconvex or concave on [ a, b ] we have from (19) f ( b ) − b − a Z ba f ( x ) dx ≤ b − a f ′ ( a ) + 2 f ′ ( b )]and f ( a ) − b − a Z ba f ( x ) dx ≤ − b − a f ′ ( a ) + f ′ ( b )] , respectively. he Hermite-Hadamard inequality revisited: Some new proofs and applications 11 The first moment of a function f about the center point c = ( a + b ) / M f = R ba (cid:0) x − a + b (cid:1) f ( x ) dx . In the following Theorem we obtainsome estimation from above and below for M f , when f ′ is convex. Theorem 7
Suppose that the derivative f ′ of the function f : [ a, b ] → R isconvex. Then the first moment of f about the center point c = ( a + b ) / satisfiesthe following inequality A ≤ Z ba (cid:18) x − a + b (cid:19) f ( x ) dx ≤ B, (24) where A = ( a − b ) f ( b ) − f ( a )) − ( b − a )
48 ( f ′ ( a ) + f ′ ( b )) and B = ( b − a )
24 ( f ′ ( a ) + f ′ ( b )) . Proof
Integration by parts leads to Z ba ( x − a )( b − x ) f ′ ( x ) dx = Z ba ( x − a )( b − x ) df ( x )= 2 Z ba (cid:18) x − a + b (cid:19) f ( x ) dx. Hence, Z ba (cid:18) x − a + b (cid:19) f ( x ) dx = 12 Z ba ( x − a )( b − x ) f ′ ( x ) dx (set x = (1 − t ) a + tb , ( x − a )( b − x ) = ( b − a ) t (1 − t ), 0 ≤ t ≤ b − a ) Z t (1 − t ) f ′ ((1 − t ) a + tb ) dt ≤
12 ( b − a ) Z t (1 − t )[ f ′ ( a )(1 − t ) + f ′ ( b ) t ] dt = ( b − a )
24 ( f ′ ( a ) + f ′ ( b )) . This proved the right hand side of (24).Further, again using integration by parts we have Z ba (cid:18) x − a + b (cid:19) f ′ ( x ) dx = ( b − a ) f ( b ) − f ( a )) − Z ba (cid:18) x − a + b (cid:19) f ( x ) dx, and therefore, Z ba (cid:18) x − a + b (cid:19) f ( x ) dx = ( b − a ) f ( b ) − f ( a )) − Z ba (cid:18) x − a + b (cid:19) f ′ ( x ) dx. (25)Furhermore, setting x = (1 − t ) a + tb , (cid:0) x − a + b (cid:1) = ( b − a ) (cid:0) t − (cid:1) and dx = ( b − a ) dt , (0 ≤ t ≤ Z ba (cid:18) x − a + b (cid:19) f ′ ( x ) dx = ( b − a ) Z (cid:18) t − (cid:19) f ′ ((1 − t ) a + tb ) dt ≤ ( b − a ) Z (cid:18) t − (cid:19) [(1 − t ) f ′ ( a ) + tf ′ ( b )] dt = ( b − a ) " f ′ ( a ) Z (cid:18) t − (cid:19) (1 − t ) dt + f ′ ( b ) Z t (cid:18) t − (cid:19) dt = ( b − a )
24 ( f ′ ( a ) + f ′ ( b )) . Taking into account this in (25) we obtain the left hand side of inequality (24).The proof is complete.A straightforward calculation shows that the equality in both sides of (24)is attained for f ( x ) = k ( x − ( a + b ) x ) + n , where k and n are arbitrary realnumbers. Theorem 8
Given f : [ a, b ] → R , let f ′′ be convex. Then the following in-equality holds A ≤ f ( a ) + f ( b )2 − b − a Z ba f ( x ) dx ≤ B, (26) where A = b − a f ′ ( b ) − f ′ ( a )) − ( b − a )
48 ( f ′′ ( a ) + f ′′ ( b )) and B = ( b − a )
24 ( f ′′ ( a ) + f ′′ ( b )) . Proof
Integration by parts twice gives Z ba ( x − a )( b − x ) f ′′ ( x ) dx = ( b − a )( f ( a ) + f ( b )) − Z ba f ( x ) dx. he Hermite-Hadamard inequality revisited: Some new proofs and applications 13 Hence, f ( a ) + f ( b )2 − b − a Z ba f ( x ) dx = 12( b − a ) Z ba ( x − a )( b − x ) f ′′ ( x ) dx (Set x = (1 − t ) a + tb , 0 ≤ t ≤ b − a ) Z t (1 − t ) f ′′ ((1 − t ) a + tb ) dt ≤ ( b − a ) Z t (1 − t )[(1 − t ) f ′′ ( a ) + tf ′′ ( b )] dt = ( b − a )
24 ( f ′′ ( a ) + f ′′ ( b )) . The right hand side of (26) is proved.Straightforward calculations show that, integration by parts twice yields Z ba (cid:18) x − a + b (cid:19) f ′′ ( x ) dx = (cid:18) b − a (cid:19) ( f ′ ( b ) − f ′ ( a )) − b − a ) " f ( a ) + f ( b )2 − b − a Z ba f ( x ) dx . Hence, f ( a ) + f ( b )2 − b − a Z ba f ( x ) dx = b − a f ′ ( b ) − f ′ ( a )) − b − a ) Z ba (cid:18) x − a + b (cid:19) f ′′ ( x ) dx. (27)Setting x = (1 − t ) a + tb , (0 ≤ t ≤
1) and using the convexity of f ′′ , we have Z ba (cid:18) x − a + b (cid:19) f ′′ ( x ) dx = ( b − a ) Z (cid:18) t − (cid:19) f ′′ ((1 − t ) a + tb ) dt ≤ ( b − a ) " f ′′ ( a ) Z (cid:18) t − (cid:19) (1 − t ) dt + f ′′ ( b ) Z (cid:18) t − (cid:19) tdt = ( b − a )
24 ( f ′′ ( a ) + f ′′ ( b )) . By making use of this in (27) we obtain the left hand side of inequality (26).The proof is complete.It is easy to verify that the equality in both sides of (24) is attained for thefunctions f ( x ) = k (2 x − a + b ) x ) + mx + n , with arbitrary real numbers k , m and n . Remark 9
There are several results in the literature under the condition of theconvexity of | f ′ | or | f ′′ | (see, e.g. [2,6,24]). As far as we know, the conditionsand assertions of our theorems 7, 8 and 9 completely differ from those knownin the literature.In the following theorem we give some estimations for the mean value of afunction f whose first derivative is convex. Theorem 9
Let f : [ a, b ] → R be differentiable and its derivative f ′ be convex.Then(a) N ≤ b − a Z ba f ( x ) dx ≤ M, (28) where N = 13 ( f ( a ) + 2 f ( b )) − f ′ ( b )( b − a ) and M = 13 ( f ( b ) + 2 f ( a )) + 16 f ′ ( a )( b − a ); (b) N ≤ b − a Z ba f ( x ) dx ≤ M , (29) where N = f ( a ) + 2 f (cid:18) a + b (cid:19) − b − a Z a + b a f ( x ) dx and M = f ( b ) + 2 f (cid:18) a + b (cid:19) − b − a Z b a + b f ( x ) dx. Corollary 4 Z b a + b f ( x ) dx − Z a + b a f ( x ) dx ≤
14 ( b − a )( f ( b ) − f ( a )) . (30) Proof
Since f ′ is convex, (HH) leads to f ′ (cid:18) a + x (cid:19) ≤ x − a ( f ( x ) − f ( a )) ≤ f ′ ( a ) + f ′ ( x )2 ; (31) f ′ (cid:18) x + b (cid:19) ≤ b − x ( f ( b ) − f ( x )) ≤ f ′ ( x ) + f ′ ( b )2 . (32)Multiplying the inequalities (31) by ( x − a ) and integrating over [ a, b ], aftersimple calculations we obtain2( b − a ) f (cid:18) a + b (cid:19) − Z a + b a f ( x ) dx ≤ Z ba f ( x ) dx − f ( a )( b − a ) he Hermite-Hadamard inequality revisited: Some new proofs and applications 15 ≤ f ′ ( a )( b − a ) + 12 ( b − a ) f ( b ) − Z ba f ( x ) dx. The above inequalities can be written as two seperate inequalities:1 b − a Z ba f ( x ) dx ≤
13 (2 f ( a ) + f ( b )) + 16 f ′ ( a )( b − a ) (33)and 1 b − a Z ba f ( x ) dx + 4 b − a Z a + b a f ( x ) dx ≥ f ( a ) + 2 f (cid:18) a + b (cid:19) . (34)In a similar way, multiplying inequalities (32) by ( b − x ) and integrating over[ a, b ], after some calculations we have the following two inequalities:1 b − a Z ba f ( x ) dx ≥
13 ( f ( a ) + 2 f ( b )) − f ′ ( b )( b − a ) (35)and 1 b − a Z ba f ( x ) dx + 4 b − a Z b a + b f ( x ) dx ≤ f ( b ) + 2 f (cid:18) a + b (cid:19) . (36)Now, the inequalities (33) and (35) yields (28) and the inequalities (34) and(36) yields (29). The Corollary follows by subtracting (34) from (36).The proof is complete. Example 3
For 0 < a < x < b < ∞ and f ( x ) = ln x , the inequality (30) yields a a + b a + b ) · b a +3 b a + b ) ≤ a + b . (37)Since α + β = 1 for α = a + b a + b ) and β = a +3 b a + b ) , then by the generalizedAM-GM inequality we have a α · b β < α · a + β · b = 3 a + b a + b ) · a + a + 3 b a + b ) · b. (38)A simple calculation shows that a + b < a + b a + b ) · a + a + 3 b a + b ) · b, and therefore, the inequality (37) is better than (38). References
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