Matrix Fejér and Levin-Stečkin Inequalities
aa r X i v : . [ m a t h . F A ] M a r MATRIX FEJ ´ER AND LEVIN-STE ˇCKIN INEQUALITIES
MOHAMMAD SABABHEH , SHIVA SHEYBANI AND HAMID REZA MORADI Abstract.
Fej´er and Levin-Steˇckin inequalities treat integrals of the product of convex func-tions with symmetric functions. The main goal of this article is to present possible matrixversions of these inequalities. In particular, majorization results are shown of Fej´er type forboth convex and log-convex functions. For matrix Levin-Steˇckin type, we present more rigorousresults involving the partial Loewner ordering for Hermitian matrices. Introduction
The theory of Convex functions has played a major role in the study of Mathematical in-equalities. Related to convex-type inequalities, the Levin-Steˇckin’s inequality states that if thefunction p : [0 , → R is symmetric about t = , namely p (1 − t ) = p ( t ) , and non-decreasingon (cid:2) , (cid:3) , then for every convex function f on [0 , Z p ( t ) f ( t ) dt ≤ Z p ( t ) dt Z f ( t ) dt holds true [7]. If p is symmetric non-negative (without any knowledge about its monotonicity)and f is convex, Fej´er inequality states that [5] Z p ( t ) dtf (cid:18) a + b (cid:19) ≤ Z p ( t ) f ((1 − t ) a + tb ) dt ≤ Z p ( t ) dt f ( a ) + f ( b )2 . We notice that Fej´er inequality reduces to the Hermite-Hadamard inequality [6], when p ( t ) = 1 . In the field of Mathematical inequalities, it is of interest to extend known inequalities from thesetting of scalars to other objects; such as matrices. In this article, we will be interested inextending both the Levin-Steˇckin and Fej´er inequalities to the matrices setting.In the sequel, M n will denote the algebra of all n × n complex matrices. The conjugatetranspose (or adjoint) of A ∈ M n is denoted by A ∗ , and then the matrix A will be calledHermitian if A ∗ = A. When h Ax, x i ≥ x ∈ C n , A is said to be positive semidefinite, andis denoted as A ≥ . If A ≥ A is invertible, then A is said to be positive (strictly positive Mathematics Subject Classification.
Primary 47A63, Secondary 47B15, 15A45, 47A30, 15A60.
Key words and phrases.
Levin-Steˇckin inequality, Fej´er inequality, positive matrices. or positive definite). When A, B ∈ M n are Hermitian, we say that A ≤ B if B − A ≥ . Thisprovides a partial ordering on the class of Hermitian matrices. The eigenvalues of a Hermitianmatrix A will be denoted by λ ( A ) , λ ( A ) , · · · , λ n ( A ), repeated according to their multiplicityand arranged decreasingly. That is λ ( A ) ≥ λ ( A ) ≥ · · · ≥ λ n ( A ) . The relation A ≤ B implies λ i ( A ) ≤ λ i ( B ) for any such Hermitian matrices A, B ∈ M n .However, the converse is not true. This urges the need to discuss, in some cases, the latterorder. For convenience, we will write λ ( A ) ≤ λ ( B ) to mean that λ i ( A ) ≤ λ ( B ) , i = 1 , , · · · , n. Another weaker ordering among matrices is the so called weak majorization ≺ w , defined forthe Hermitian matrices A, B as A ≺ w B ⇔ k X i =1 λ i ( A ) ≤ k X i =1 λ i ( B ) , k = 1 , · · · , n. It is clear that [1] A ≤ B ⇒ λ ( A ) ≤ λ ( B ) ⇒ A ≺ w B. It is customary to obtain one of these orders when extending a scalar inequality to a matrixinequality. For example, in this article we obtain λ Z p ( t ) dtf (cid:18) A + B (cid:19) ≤ λ Z p ( t ) f ((1 − t ) A + tB ) dt , as an extension of Fej´er inequality, to the Hermitian matrices A, B with spectra in the domainof f .Further, if f is monotone, then λ Z p ( t ) f ((1 − t ) A + tB ) dt ≤ λ Z p ( t ) dt f ( A ) + f ( B )2 ;as matrix inequalities of the Fej´er inequality.In the next section we study the possible matrix versions of Fej´er inequality, which in turnsimply certain versions of the Hermite-Hadamard matrix inequality [8]. Then log-convex func-tions will be deployed to obtain new matrix Fej´er inequalities for this type of functions, and weconclude with the discussion of the matrix Levin-Steˇckin inequality.2. Fej´er matrix inequalities for convex functions
We begin with the following weak majorization of Fej´er-type inequality.
Theorem 2.1.
Let f : J → R be convex and let p : [0 , → [0 , ∞ ) be symmetric about t = . If A, B ∈ M n are Hermitian with spectra in the interval J , then λ Z p ( t ) dtf (cid:18) A + B (cid:19) ≺ w λ Z p ( t ) f ((1 − t ) A + tB ) dt . Proof. If f is a convex function, then for any 0 ≤ t ≤ f (cid:18) a + b (cid:19) = f (cid:18) (1 − t ) a + tb + (1 − t ) b + ta (cid:19) ≤ f ((1 − t ) a + tb ) + f ((1 − t ) b + ta )2 . Thus,(2.1) f (cid:18) a + b (cid:19) ≤ f ((1 − t ) a + tb ) + f ((1 − t ) b + ta )2 . If the function p : [0 , → R is symmetric, we get from (2.1), p ( t ) f (cid:18) a + b (cid:19) ≤ p ( t ) (cid:18) f ((1 − t ) a + tb ) + f ((1 − t ) b + ta )2 (cid:19) . Integrating on [0 , Z p ( t ) dtf (cid:18) a + b (cid:19) ≤ Z p ( t ) f ((1 − t ) a + tb ) dt. If we replace a , b by h Ax, x i , h Bx, x i respectively, in (2.2), we get(2.3) Z p ( t ) dtf (cid:18) h Ax, x i + h Bx, x i (cid:19) ≤ Z p ( t ) f ((1 − t ) h Ax, x i + t h Bx, x i ) dt. On the other hand, it follows from the Jensen’s inequality [9, Theorem 1.2], f ( h ((1 − t ) A + tB ) x, x i ) ≤ h f ((1 − t ) A + tB ) x, x i . By multiplying both sides by p ( t ), we get p ( t ) f ( h ((1 − t ) A + tB ) x, x i ) ≤ p ( t ) h f ((1 − t ) A + tB ) x, x i . Therefore,(2.4) Z p ( t ) f ( h ((1 − t ) A + tB ) x, x i ) dt ≤ * Z p ( t ) f ((1 − t ) A + tB ) dt x, x + . Combining inequalities (2.3) with (2.4), we obtain(2.5) Z p ( t ) dtf (cid:18) h Ax, x i + h Bx, x i (cid:19) ≤ * Z p ( t ) f ((1 − t ) A + tB ) dt x, x + . Suppose that λ , . . . , λ m are the eigenvalues of A + B with x , . . . , x m as an orthonormal systemof corresponding eigenvectors arranged such that f ( λ ) ≥ · · · ≥ f ( λ m ). We have k X j =1 λ j Z p ( t ) dtf (cid:18) A + B (cid:19) = k X j =1 1 Z p ( t ) dtf (cid:18)(cid:28)(cid:18) A + B (cid:19) x j , x j (cid:29)(cid:19) = k X j =1 1 Z p ( t ) dtf (cid:18) h Ax j , x j i + h Bx j , x j i (cid:19) ≤ k X j =1 * Z p ( t ) f ((1 − t ) A + tB ) dt x j , x j + (by the inequality (2.5)) ≤ k X j =1 λ j Z p ( t ) f ((1 − t ) A + tB ) dt . Namely, k X j =1 λ j Z p ( t ) dtf (cid:18) A + B (cid:19) ≤ k X j =1 λ j Z p ( t ) f ((1 − t ) A + tB ) dt . Therefore, λ Z p ( t ) dtf (cid:18) A + B (cid:19) ≺ w λ Z p ( t ) f ((1 − t ) A + tB ) dt . (cid:3) Fej´er inequalities via log-convex functions
In this part of the paper, we show a matrix Fej´er inequality for log-convex functions.
Theorem 3.1.
Let f be log-convex and p : [0 , → (0 , ∞ ) be symmetric and normalized in thesense that R p ( t ) dt = 1 . Then λ (cid:18) log f (cid:18) A + B (cid:19)(cid:19) ≺ w λ (cid:18) log Z p ( t ) f ((1 − t ) A + tB ) dt (cid:19) . Proof.
When f is convex, we have f (cid:18)(cid:28) A + B x, x (cid:29)(cid:19) ≤ Z p ( t ) f h ((1 − t ) A + tB ) x, x i dt, for any unit vector x . Since f is log-convex, it follows thatlog f (cid:18)(cid:28) A + B x, x (cid:29)(cid:19) ≤ Z p ( t ) log f h ((1 − t ) A + tB ) x, x i dt. Noting that log is a concave function and that dµ ( t ) := p ( t ) dt is a probability measure, we havelog f (cid:18)(cid:28) A + B x, x (cid:29)(cid:19) ≤ Z p ( t ) log f h ((1 − t ) A + tB ) x, x i dt = Z log f h ((1 − t ) A + tB ) x, x i dµ ( t ) ≤ log Z f h ((1 − t ) A + tB ) x, x i dµ ( t )= log Z p ( t ) f h ((1 − t ) A + tB ) x, x i dt, for any unit vector x . Now let λ j be the eigenvalues of A + B with orthonormal eigenvectors x , x , · · · , x n , so that f ( λ ) ≥ · · · ≥ f ( λ n ) . Then k X j =1 λ j (cid:18) log f (cid:18) A + B (cid:19)(cid:19) = k X j =1 log f ( λ j )= k X j =1 log f (cid:18)(cid:28) A + B x j , x j (cid:29)(cid:19) ≤ k X j =1 log Z p ( t ) f h ((1 − t ) A + tB ) x j , x j i dt ≤ k X j =1 λ j (cid:18) log Z p ( t ) f ((1 − t ) A + tB )) dt (cid:19) . This completes the proof. (cid:3)
As a consequence, we have the following.
Corollary 3.1.
Let f be log-convex and p : [0 , → (0 , ∞ ) be symmetric and normalized. Then k Y j =1 λ j (cid:18) f (cid:18) A + B (cid:19)(cid:19) ≤ k Y j =1 λ j (cid:18)Z p ( t ) f ((1 − t ) A + tB ) dt (cid:19) , k = 1 , · · · , n. Proof.
From Theorem3.1, we have k X j =1 λ j (cid:18) log f (cid:18) A + B (cid:19)(cid:19) ≤ k X j =1 λ j (cid:18) log Z p ( t ) f ((1 − t ) A + tB )) dt (cid:19) ⇒ k X j =1 log λ j (cid:18) f (cid:18) A + B (cid:19)(cid:19) ≤ k X j =1 log λ j (cid:18)Z p ( t ) f ((1 − t ) A + tB )) dt (cid:19) ⇒ log k Y j =1 λ j (cid:18) f (cid:18) A + B (cid:19)(cid:19) ≤ log k Y j =1 λ j (cid:18)Z p ( t ) f ((1 − t ) A + tB ) dt (cid:19) , which implies the desired inequality. (cid:3) Levin-Steˇckin matrix inequalities
We begin by presenting a new inequality of Levin-Steˇckin type. The significance of thisinequality is its validity for any positive function p ; without imposing any conditions on itssymmetry or monotony. Theorem 4.1.
Let f : [0 , → R be convex differentiable and let p : [0 , → [0 , ∞ ) becontinuous. Then Z f ( t ) dt Z p ( t ) dt + (cid:18)Z f ′ ( t ) dt Z tp ( t ) dt − Z tf ′ ( t ) dt Z p ( t ) dt (cid:19) ≤ Z f ( t ) p ( t ) dt. Further, Z p ( t ) f ( t ) dt + 12 Z p ( t ) f ′ ( t ) dt − Z p ( t ) tf ′ ( t ) dt ≤ Z p ( t ) dt Z f ( t ) dt. Proof.
For the convex differentiable function f and s, t ∈ [0 ,
1] we have(4.1) f ( s ) + f ′ ( s )( t − s ) ≤ f ( t ) . Since p ( t ) ≥ , it follows that p ( t ) f ( s ) + p ( t ) f ′ ( s )( t − s ) ≤ p ( t ) f ( t ) , s, t ∈ [0 , . Integrating this inequality over t ∈ [0 ,
1] then over s ∈ [0 ,
1] implies Z f ( s ) dt Z p ( t ) dt + (cid:18)Z f ′ ( s ) dt Z tp ( t ) dt − Z sf ′ ( s ) dt Z p ( t ) dt (cid:19) ≤ Z f ( t ) p ( t ) dt, which is equivalent to the first desired inequality.For the second inequality, integrating (4.1) over t ∈ [0 , f ( s ) + f ′ ( s ) (cid:18) − s (cid:19) ≤ Z f ( t ) dt. If we put s = t , we have f ( t ) + f ′ ( t ) (cid:18) − t (cid:19) ≤ Z f ( t ) dt. Multiplying both sides by p ( t ), we get p ( t ) f ( t ) + p ( t ) f ′ ( t ) (cid:18) − t (cid:19) ≤ p ( t ) Z f ( t ) dt. Again, if we take integral over t ∈ [0 ,
1] we infer that Z p ( t ) f ( t ) dt + 12 Z p ( t ) f ′ ( t ) dt − Z p ( t ) tf ′ ( t ) dt ≤ Z p ( t ) dt Z f ( t ) dt. This completes the proof. (cid:3)
Corollary 4.1.
Let f : [0 , → R be convex differentiable and let p : [0 , → [0 , ∞ ) besymmetric about and non-decreasing on (cid:2) , (cid:3) . Then Z f ′ ( t ) dt Z tp ( t ) dt ≤ Z tf ′ ( t ) dt Z p ( t ) dt. Proof.
This follows from the first inequality in Theorem 4.1 because when p is symmetric about1 / (cid:2) , (cid:3) , we have Z f ( t ) p ( t ) dt ≤ Z f ( t ) dt Z p ( t ) dt. (cid:3) The following is the operator Levin-Stˇeckin inequality, see also [4, Theorem 2].
Theorem 4.2.
Let f be operator convex and p : [0 , → [0 , ∞ ) be symmetric about t = andnon-decreasing on (cid:2) , (cid:3) . Then Z p ( t ) f ((1 − t ) A + tB ) dt ≤ Z p ( t ) dt Z f ((1 − t ) A + tB ) dt. Proof.
Let x ∈ C n be a unit vector. Since the function F ( t ) = h f ((1 − t ) A + tB ) x, x i is areal-valued convex function on [0 ,
1] (see [3, Theorem 1]), we have * Z p ( t ) f ((1 − t ) A + tB ) dtx, x + = Z p ( t ) h f ((1 − t ) A + tB ) x, x i dt ≤ Z p ( t ) dt Z h f ((1 − t ) A + tB ) x, x i dt = * Z p ( t ) dt Z f ((1 − t ) A + tB ) dtx, x + . Therefore, Z p ( t ) f ((1 − t ) A + tB ) dt ≤ Z p ( t ) dt Z f ((1 − t ) A + tB ) dt. (cid:3) The following theorem gives a reverse for the operator Levin-Stˇeckin’s inequality by employingthe Mond- Peˇcari´c method [9].
Theorem 4.3.
Let f : [ m, M ] → R be convex and let p : [0 , → [0 , ∞ ) be symmetric about t = . If A, B ∈ M n are Hermitian with spectra in the interval [ m, M ] , then for any α ≥ Z p ( t ) dt Z f ((1 − t ) A + tB ) dt ≤ β Z p ( t ) dtI + α Z p ( t ) f ((1 − t ) A + tB ) dt, where β = max m ≤ x ≤ M { a f x + b f − αf ( x ) } , a f = f ( M ) − f ( m ) M − m , and a f = Mf ( m ) − mf ( M ) M − m .Proof. Since f ( x ) ≤ a f x + b f , we get by the functional calculus f ((1 − t ) A + tB ) ≤ a f ((1 − t ) A + tB ) + b f I. By taking integral over 0 ≤ t ≤ Z f ((1 − t ) A + tB ) dt ≤ a f (cid:18) A + B (cid:19) + b f I. This implies, Z p ( t ) dt Z f ((1 − t ) A + tB ) dt ≤ Z p ( t ) dta f (cid:18) A + B (cid:19) + Z p ( t ) dtb f I. Hence for any vector x , * Z p ( t ) dt Z f ((1 − t ) A + tB ) dt x, x + ≤ Z p ( t ) dta f (cid:28)(cid:18) A + B (cid:19) x, x (cid:29) + Z p ( t ) dtb f . Now, by (2.3), we can write * Z p ( t ) dt Z f ((1 − t ) A + tB ) dt x, x + − α Z p ( t ) f ( h ((1 − t ) A + tB ) x, x i ) dt ≤ Z p ( t ) dta f (cid:28)(cid:18) A + B (cid:19) x, x (cid:29) + Z p ( t ) dtb f − α Z p ( t ) f ( h ((1 − t ) A + tB ) x, x i ) dt ≤ Z p ( t ) dta f (cid:28)(cid:18) A + B (cid:19) x, x (cid:29) + Z p ( t ) dtb f − α Z p ( t ) dtf (cid:18)(cid:28)(cid:18) A + B (cid:19) x, x (cid:29)(cid:19) = Z p ( t ) dt (cid:18) a f (cid:28)(cid:18) A + B (cid:19) x, x (cid:29) + b f − αf (cid:18)(cid:28)(cid:18) A + B (cid:19) x, x (cid:29)(cid:19)(cid:19) ≤ Z p ( t ) dt max m ≤ x ≤ M { a f x + b f − αf ( x ) } . Thus, * Z p ( t ) dt Z f ((1 − t ) A + tB ) dt x, x + ≤ Z p ( t ) dtβ + α Z p ( t ) f ( h ((1 − t ) A + tB ) x, x i ) dt ≤ Z p ( t ) dtβ + α Z p ( t ) h f ((1 − t ) A + tB ) x, x i dt (by [9, Theorem 1.2])= * Z p ( t ) dtβI + α Z p ( t ) f ((1 − t ) A + tB ) dt x, x + as desired. (cid:3) Further inequalities via synchronous functions
We say that the functions f, g : J → R are synchronous (asynchronous) on the interval J ifthey satisfy the following condition:( f ( t ) − f ( s )) ( g ( t ) − g ( s )) ≥ ( ≤ ) 0 , ∀ s, t ∈ J. It is obvious that, if f, g are monotonic and have the same monotonicity on the interval J , thenthey are synchronous on J while if they have opposite monotonicity, they are asynchronous. Related to the Levin-Steˇckin inequality, the celebrated ˆCebyˆsev inequality [2] states that if f and g are two functions having the same monotonicity on [0 , Z f ( t ) dt Z g ( t ) dt ≤ Z f ( t ) g ( t ) dt. The following result provides a refinement and a reverse of this inequality, via synchronousfunctions.
Theorem 5.1.
Let f, g : [ a, b ] → R be synchronous functions on the interval [ a, b ] . Then min b − a b Z a f ( t ) dt − b − a b Z a f ( t ) dt , b − a b Z a g ( t ) dt − b − a b Z a g ( t ) dt ≤ b − a b Z a f ( t ) g ( t ) dt − b − a b Z a f ( t ) dt b − a b Z a g ( t ) dt max b − a b Z a f ( t ) dt − b − a b Z a f ( t ) dt , b − a b Z a g ( t ) dt − b − a b Z a g ( t ) dt . If f and g have the opposite monotonicity then min b − a b Z a f ( t ) dt − b − a b Z a f ( t ) dt , b − a b Z a g ( t ) dt − b − a b Z a g ( t ) dt ≤ b − a b Z a f ( t ) dt b − a b Z a g ( t ) dt − b − a b Z a f ( t ) g ( t ) dt ≤ max b − a b Z a f ( t ) dt − b − a b Z a f ( t ) dt , b − a b Z a g ( t ) dt − b − a b Z a g ( t ) dt . Proof.
We prove the first inequality. The second inequality goes likewise and we omit thedetails. We have f ( t ) g ( t ) + f ( s ) g ( s ) − ( f ( t ) g ( s ) + f ( s ) g ( t ))= ( f ( t ) − f ( s )) ( g ( t ) − g ( s ))= | ( f ( t ) − f ( s )) ( g ( t ) − g ( s )) | = | f ( t ) − f ( s ) | | g ( t ) − g ( s ) |≥ min (cid:8) ( f ( t ) − f ( s )) , ( g ( t ) − g ( s )) (cid:9) = min (cid:8) f ( t ) + f ( s ) − f ( t ) f ( s ) , g ( t ) + g ( s ) − g ( t ) g ( s ) (cid:9) . Therefore, min (cid:8) f ( s ) + f ( t ) − f ( s ) f ( t ) , g ( t ) + g ( s ) − g ( t ) g ( s ) (cid:9) ≤ f ( t ) g ( t ) + f ( s ) g ( s ) − ( f ( t ) g ( s ) + f ( s ) g ( t )) . Consequently,min ( b − a ) f ( s ) + b Z a f ( t ) dt − f ( s ) b Z a f ( t ) dt, b Z a g ( t ) dt + ( b − a ) g ( s ) − g ( s ) b Z a g ( t ) dt ≤ b Z a f ( t ) g ( t ) dt + ( b − a ) f ( s ) g ( s ) − g ( s ) b Z a f ( t ) dt − f ( s ) b Z a g ( t ) dt. Upon integration, this impliesmin b − a ) b Z a f ( t ) dt − b Z a f ( t ) dt , b − a ) b Z a g ( t ) dt − b Z a g ( t ) dt ≤ b − a ) b Z a f ( t ) g ( t ) dt − b Z a f ( t ) dt b Z a g ( t ) dt. Multiplying both sides by 1 / b − a ) , we obtain,min b − a b Z a f ( t ) dt − b − a b Z a f ( t ) dt , b − a b Z a g ( t ) dt − b − a b Z a g ( t ) dt ≤ b − a b Z a f ( t ) g ( t ) dt − b − a b Z a f ( t ) dt b − a b Z a g ( t ) dt. The second inequality obtains from the same arguments and the following relationmax (cid:8) f ( s ) + f ( t ) − f ( s ) f ( t ) , g ( t ) + g ( s ) − g ( t ) g ( s ) (cid:9) ≥ f ( t ) g ( t ) + f ( s ) g ( s ) − ( f ( t ) g ( s ) + f ( s ) g ( t )) . (cid:3) We establish a refinement and a reverse for the Levin-Steˇckin inequality in the next result.
Theorem 5.2.
Let p : [0 , → R be a symmetric about t = , namely p (1 − t ) = p ( t ) , andnon-decreasing on (cid:2) , (cid:3) , then for every convex function f on [0 , , Z p ( t ) f ( t ) dt ≤ Z p ( t ) dt Z f ( t ) dt − min / Z p ( t ) dt − Z p ( t ) dt , / Z ( f ( t ) + f (1 − t )) dt − Z ( f ( t ) + f (1 − t )) dt . A similar but reversed inequality holds if we replace min with max .Proof.
If if f is symmetric and convex, by Theorem 5.1, we have Z p ( t ) dt Z f ( t ) dt = / Z p ( t ) dt + Z / p ( t ) dt / Z f ( t ) dt + Z / f ( t ) dt = 4 / Z p ( t ) dt / Z f ( t ) dt ≥ / Z p ( t ) f ( t ) dt + min / Z p ( t ) dt − / Z p ( t ) dt , / Z f ( t ) dt − / Z f ( t ) dt = Z p ( t ) f ( t ) dt + min / Z p ( t ) dt − Z p ( t ) dt , / Z f ( t ) dt − Z f ( t ) dt . Namely, Z p ( t ) f ( t ) dt + min / Z p ( t ) dt − Z p ( t ) dt , / Z f ( t ) dt − Z f ( t ) dt ≤ Z p ( t ) dt Z f ( t ) dt. We shall consider now an arbitrary f . For convex f the function f ( x )+ f (1 − x )2 is convex andsymmetric, so we can use the above inequality. Hence, Z p ( t ) f ( t ) dt = R p ( t ) f ( t ) dt + R p (1 − t ) f (1 − t ) dt Z p ( t ) f ( t ) + f (1 − t )2 dt ≤ Z p ( t ) dt Z f ( t ) + f (1 − t )2 dt − min / Z p ( t ) dt − Z p ( t ) dt , / Z ( f ( t ) + f (1 − t )) dt − Z ( f ( t ) + f (1 − t )) dt = Z p ( t ) dt Z f ( t ) dt − min / Z p ( t ) dt − Z p ( t ) dt , / Z ( f ( t ) + f (1 − t )) dt − Z ( f ( t ) + f (1 − t )) dt , which yields the desired inequality. (cid:3) We can improve the second inequality in Theorem 5.1, in the following way. Theorem 5.3.
Let f, g : J → R be synchronous functions on the interval [0 , . Then Z f ( t ) g ( t ) dt − Z f ( t ) dt Z g ( t ) dt ≤ Z f ( t ) dt − Z f ( t ) dt + Z g ( t ) dt − Z g ( t ) dt . Proof.
We have f ( t ) g ( t ) + f ( s ) g ( s ) − ( f ( t ) g ( s ) + f ( s ) g ( t ))= ( f ( t ) − f ( s )) ( g ( t ) − g ( s ))= | ( f ( t ) − f ( s )) ( g ( t ) − g ( s )) | = | f ( t ) − f ( s ) | | g ( t ) − g ( s ) |≤ (cid:0) ( f ( t ) − f ( s )) + ( g ( t ) − g ( s )) (cid:1) = 12 (cid:0) f ( t ) + f ( s ) + g ( t ) + g ( s ) − g ( t ) g ( s ) + f ( t ) f ( s )) (cid:1) . Therefore, f ( t ) g ( t ) + f ( s ) g ( s ) − ( f ( t ) g ( s ) + f ( s ) g ( t )) ≤ (cid:0) f ( t ) + f ( s ) + g ( t ) + g ( s ) − g ( t ) g ( s ) + f ( t ) f ( s )) (cid:1) . The remaining part of the proof is similar to the proof of Theorem 5.1, so we omit the details (cid:3)
Acknowledgement
The authors would like to thank Prof. J. C. Bourin, who pointed out a crucial mistake inthe previous version of the manuscript.
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E-mail address: [email protected] Department of Mathematics, Islamic Azad University, Mashhad Branch, Mashhad, Iran
E-mail address: [email protected] Department of Mathematics, Payame Noor University (PNU), P.O. Box 19395-4697, Tehran, Iran