A discrete inequality of Grüss type in normed linear sapces and applications for the Fourier transform, Mellin transform of sequences, for polynomials with coefficients in normed spaces and for vector valued Lipschitzian mappings, are given

Dragomir, Sever S

English

A discrete inequality of Grüss type in normed linear spaces and applications for the discrete Fourier transform, Mellin transform of sequences, for polynomials with coefficients in normed spaces and for vector valued Lipschitzian mappings are given

Victoria University Institutional Repository (VUIR)

A Grüss Type Inequality for Sequences of Vectors in Normed Linear Spaces and Applications

Another Grüss Type Inequality for Sequences of Vectors in Normed Linear Spaces and Applications

Victoria University Eprints Repository

ANOTHER GRU¨SS TYPE INEQUALITY FOR SEQUENCES OFVECTORS IN NORMED LINEAR SPACES AND APPLICATIONSS.S. DRAGOMIRAbstract. A discrete inequality of Gru¨ss type in normed linear sapces andapplications for the Fourier transform, Mellin transform of sequences, for poly-nomials with coefficients in normed spaces and for vector valued Lipschitzianmappings, are given.1. IntroductionIn 1935, G. Gru¨ss [9] proved the following integral inequality which gives anapproximation of the integral of the product in terms of the product of the integralsas follows ∣∣∣∣∣ 1b− a∫ baf (x) g (x) dx− 1b− a∫ baf (x) dx · 1b− a∫ bag (x) dx∣∣∣∣∣(1.1)≤ 14(Φ− φ) (Γ− γ) ,where f , g : [a, b]→ R are integrable on [a, b] and satisfy the condition(1.2) φ ≤ f (x) ≤ Φ, γ ≤ g (x) ≤ Γfor each x ∈ [a, b] , where φ,Φ, γ,Γ are given real constants.Moreover, the constant 14 is sharp in the sense that it cannot be replaced by asmaller one.For a simple proof of (1.1) as well as for some other integral inequalities of Gru¨sstype, see Chapter X of the recent book [11] and the papers [1]-[8] and [10].In 1950, M. Biernacki, H. Pidek and C. Ryll-Nardjewski [11, Chapter X] estab-lished the following discrete version of Gru¨ss’ inequality:Theorem 1. Let a = (a1, ..., an) , b = (b1, ..., bn) be two n−tuples of real numberssuch that r ≤ ai ≤ R and s ≤ bi ≤ S for i = 1, ..., n. Then one has(1.3)∣∣∣∣∣ 1nn∑i=1aibi − 1nn∑i=1ai · 1nn∑i=1bi∣∣∣∣∣ ≤ 1n [n2 ](1− 1n[n2])(R− r) (S − s) ,where [x] denotes the integer part of x, x ∈ R.A weighted version of the discrete Gru¨ss inequality was proved by J. E. Pecˇaric´in 1979 [11, Chapter X]:Date: September 16, 1999.1991 Mathematics Subject Classification. Primary26D15, 26D99; Secondary 46Bxx.Key words and phrases. Discrete Gru¨ss inequality, Normed Linear Spaces, Fourier Transform,Mellin Transform, Polynomials.12 S.S. DRAGOMIRTheorem 2. Let a and b be two monotonic n−tuples and p a positive one. Then∣∣∣∣∣ 1Pnn∑i=1piaibi − 1Pnn∑i=1piai · 1Pnn∑i=1pibi∣∣∣∣∣(1.4)≤ |an − a1| |bn − b1| max1≤k≤n−1[PkP¯k+1P 2n],where Pn :=∑ni=1 pi, and P¯k+1 = Pn − Pk+1.In 1981, A. Lupas¸, [11, Chapter X] proved some similar results for the firstdifference of a as follows.Theorem 3. Let a, b be two monotonic n−tuples in the same sense and p a positiven-tuple. Thenmin1≤i≤n−1|∆ai| min1≤i≤n−1|∆bi| 1Pnn∑i=1i2pi −(1Pnn∑i=1ipi)2(1.5)≤ 1Pnn∑i=1piaibi − 1Pnn∑i=1piai · 1Pnn∑i=1pibi≤ max1≤i≤n−1|∆ai| max1≤i≤n−1|∆bi| 1Pnn∑i=1i2pi −(1Pnn∑i=1ipi)2 ,where ∆ai := ai+1 − ai is the forward first difference. If there exist the numbersa¯, a¯1, r, r1 (rr1 > 0) such that ak = a¯ + kr and bk = a¯1 + kr1, then equality holdsin (1.5).In the recent paper [6], the authors obtained the following related resultTheorem 4. Let (X, ‖·‖) be a normed linear space over K, K = R,C, xi ∈ X,αi ∈ K and pi ≥ 0 (i = 1, ..., n) such that∑ni=1 pi = 1.Then we have the inequality∥∥∥∥∥n∑i=1piαixi −n∑i=1piαin∑i=1pixi∥∥∥∥∥(1.6)≤ max1≤j≤n−1|∆αj | max1≤j≤n−1‖∆xj‖ n∑i=1i2pi −(n∑i=1ipi)2 .The inequality (1.6) is sharp in the sense that the constant c = 1 in the right handside cannot be replaced by a smaller one.In this paper we point out another inequality of Gru¨ss type and apply it inapproximating the discrete Fourier transform, the Mellin transform of sequences,for polynomials with coefficients in normed linear spaces and for vector valuedLipschitzian mappings.2. An Inequality of Gru¨ss TypeThe following result holds.ANOTHER GRU¨SS TYPE INEQUALITY 3Theorem 5. Let (X, ‖·‖) be a normed linear space over K, K = R,C, xi ∈ X, αi∈ K and pi ≥ 0 (i = 1, ..., n) (n ≥ 2) such that∑ni=1 pi = 1. Then we have theinequality ∥∥∥∥∥n∑i=1piαixi −n∑i=1piαin∑i=1pixi∥∥∥∥∥(2.1)≤∑1≤j<i≤n(i− j) pipj(n−1∑k=1|∆αk|p) 1p(n−1∑k=1‖∆xk‖q) 1q,where p > 1, 1p +1q = 1.The constant c = 1 in the right hand side of (2.1) is sharp in the sense that itcannot be replaced by a smaller one.Proof. It is well known that the following identity holds in normed linear spaces [6]n∑i=1piαixi −n∑i=1piαin∑i=1pixi =12n∑i,j=1pipj (αi − αj) (xi − xj)=∑1≤j<i≤npipj (αi − αj) (xi − xj) .We observe that for i > j, we can writeαi − αj =i−1∑k=j∆αk, xi − xj =i−1∑k=j∆xkand by the generalized triangle inequality we get∥∥∥∥∥n∑i=1piαixi −n∑i=1piαin∑i=1pixi∥∥∥∥∥(2.2)≤∑1≤j<i≤npipj∣∣∣∣∣∣i−1∑k=j∆αk∣∣∣∣∣∣∥∥∥∥∥∥i−1∑k=j∆xk∥∥∥∥∥∥≤∑1≤j<i≤npipji−1∑k=j|∆αk|i−1∑k=j‖∆xk‖ =: A.Using Ho¨lder’s discrete inequality, we can state thati−1∑k=j|∆αk| ≤ (i− j)1qi−1∑k=j|∆αk|p 1pandi−1∑k=j‖∆xk‖ ≤ (i− j)1pi−1∑k=j‖∆xk‖q 1qwhere p, q > 1 and 1p +1q = 1, and then(2.3) A ≤∑1≤j<i≤npipj (i− j)i−1∑k=j|∆αk|p 1p i−1∑k=j‖∆xk‖q 1q .4 S.S. DRAGOMIRAsi−1∑k=j|∆αk|p ≤n−1∑k=1|∆αk|pandi−1∑k=j‖∆xk‖q ≤n−1∑k=1‖∆xk‖q ,for all 1 ≤ j < i ≤ n, then by (2.3) and (2.2) , we get the desired inequality (2.1) .To prove the sharpness of the constant, let us assume that (2.1) holds with aconstant c > 0. That is,∥∥∥∥∥n∑i=1piαixi −n∑i=1piαin∑i=1pixi∥∥∥∥∥(2.4)≤ c∑1≤j<i≤n(i− j) pipj(n−1∑k=1|∆αk|p) 1p(n−1∑k=1‖∆xk‖q) 1q.Choose n = 2. Then∥∥∥∥∥2∑i=1piαixi −2∑i=1piαi2∑i=1pixi∥∥∥∥∥ =∥∥∥∥∥∥122∑i,j=1pipj (αi − αj) (xi − xj)∥∥∥∥∥∥=∥∥∥∥∥∥∑1≤j<i≤2pipj (αi − αj) (xi − xj)∥∥∥∥∥∥= p1p2 |α1 − α2| ‖x1 − x2‖and∑1≤j<i≤2(i− j) pipj(1∑k=1|∆αk|p) 1p(1∑k=1‖∆xk‖q) 1q= p1p2 |α1 − α2| ‖x1 − x2‖ .Therefore, from (2.4) , we obtainp1p2 |α1 − α2| ‖x1 − x2‖ ≤ cp1p2 |α1 − α2| ‖x1 − x2‖for all α1 6= α2, x1 6= x2, and then c ≥ 1, which proves the sharpness of theconstant.Remark 1. A coarser upper bound, which can be more useful may be obtained byapplying Cauchy-Schwartz’s inequality:∑1≤j<i≤n(i− j) pipj ≤ ∑1≤j<i≤npipj 12  ∑1≤j<i≤npipj (i− j)2 12ANOTHER GRU¨SS TYPE INEQUALITY 5and taking into account that∑1≤j<i≤npipj =12 n∑i,j=1pipj −n∑i=jpipj=12(1−n∑i=1p2i)=12n∑i=1pi (1− pi)and∑1≤j<i≤npipj (i− j)2 = 12n∑i,j=1pipj (i− j)2= n∑i=1pin∑i=1i2pi −(n∑i=1ipi)2= n∑i=1i2pi −(n∑i=1ipi)2 .Thus, from (2.1) , we can state the inequality∥∥∥∥∥n∑i=1piαixi −n∑i=1piαin∑i=1pixi∥∥∥∥∥(2.5)≤√22[n∑i=1pi (1− pi)] 12 n∑i=1i2pi −(n∑i=1ipi)2 12×(n−1∑k=1|∆αk|p) 1p(n−1∑k=1‖∆xk‖q) 1q.The following corollary holds.Corollary 1. With the above assumptions, we have(2.6)∥∥∥∥∥ 1nn∑i=1αixi − 1nn∑i=1αi · 1nn∑i=1xi∥∥∥∥∥ ≤ n2 − 16n(n−1∑k=1|∆αk|p) 1p(n−1∑k=1‖∆xk‖q) 1q.The constant 16 is the best possible.6 S.S. DRAGOMIRProof. The proof follows by (2.1) , putting pi = 1n and taking into account that∑1≤j<i≤n(i− j)=∑1≤j≤2(2− j) +∑1≤j≤3(3− j) + ...+∑1≤j≤n(n− j)= 2 · 2− (1 + 2) + 3 · 3− (1 + 2 + 3) + ...+ n · n− (1 + 2 + ...+ n)= 12 + 22 + ...+ n2 − 1− (1 + 2)− (1 + 2 + 3)− ...− (1 + 2 + ...+ n)=n∑k=1k2 −n∑k=1k (k + 1)2=12(n∑k=1k2 −n∑k=1k)=n(n2 − 1)6,and the corollary is thus proved.Remark 2. If in (2.1) and (2.6) we assume that p = q = 2, then we get theinequalities: ∥∥∥∥∥n∑i=1piαixi −n∑i=1piαin∑i=1pixi∥∥∥∥∥(2.7)≤∑1≤i<j≤n(i− j) pipj(n−1∑k=1|∆αk|2) 12(n−1∑k=1‖∆αk‖2) 12and ∥∥∥∥∥ 1nn∑i=1αixi − 1nn∑i=1αi · 1nn∑i=1xi∥∥∥∥∥(2.8)≤ n2 − 16n(n−1∑k=1|∆αk|2) 12(n−1∑k=1‖∆xk‖2) 12respectively.3. Applications for the Discrete Fourier TransformLet (X, ‖·‖) be a normed linear space over K, K = C,R, and x¯ = (x1, ..., xn) bea sequence of vectors in X.For a given w ∈ R, define the discrete Fourier transform(3.1) Fw (x¯) (m) :=n∑k=1xk exp (2wimk) , m = 1, ..., n.The following approximation result for the Fourier transform (3.1) holds.Theorem 6. Let (X, ‖·‖) and x¯ ∈ Xn be as above. Then we have the inequality∥∥∥∥∥Fw (x¯) (m)− sin (wmn)sin (wm) exp [(n+ 1)mi]× 1nn∑k=1xk∥∥∥∥∥(3.2)≤ n+ 13(n− 1)1+ 1p |sin (wm)|(n−1∑k=1‖∆xk‖q) 1q,where p > 1, 1p +1q = 1 and m ∈ {1, ..., n} , w 6= knpi, k ∈ Z.ANOTHER GRU¨SS TYPE INEQUALITY 7Proof. Using the inequality (2.6) , we can state that∥∥∥∥∥n∑k=1akxk −n∑k=1ak · 1nn∑k=1xk∥∥∥∥∥(3.3)≤ n2 − 16(n−1∑k=1|∆ak|p) 1p(n−1∑k=1‖∆xk‖q) 1qfor all ak ∈ K, xk ∈ X, k = 1, ..., n.Now, choose in (3.3) , ak = exp (2wimk) to obtain∥∥∥∥∥Fw (x¯) (m)−n∑k=1exp (2wimk)× 1nn∑k=1xk∥∥∥∥∥(3.4)≤ n2 − 16(n−1∑k=1|exp (2wim (k + 1))− exp (2wimk)|p) 1p(n−1∑k=1‖∆xk‖q) 1q,for all m ∈ {1, ..., n} .However,n∑k=1exp (2wimk) = exp (2wim)×[exp (2wimn)− 1exp (2wim)− 1]= exp (2wim)×[cos (2wmn) + i sin (2wmn)− 1cos (2wm) + i sin (2wm)− 1]= exp (2wim)×[−2 sin2 (wmn) + 2i sin (wmn) cos (wmn)−2 sin2 (wm) + 2i sin (wm) cos (wm)]= exp (2wim)× sin (wmn)sin (wm)[sin (wmn)− i cos (wmn)sin (wm)− i cos (wm)]= exp (2wim)× sin (wmn)sin (wm)[cos (wmn) + i sin (wmn)cos (wm) + i sin (wm)]=sin (wmn)sin (wm)× exp (2wim)[exp (iwmn)exp (iwm)]=sin (wmn)sin (wm)× exp [2wim+ iwmn− iwm]=sin (wmn)sin (wm)× exp [(n+ 1)mi] .We observe thatexp (2wim (k + 1))− exp (2wimk)= cos (2wm (k + 1)) + i sin (2wm (k + 1))− cos (2wmk)− i sin (2wmk)= cos (2wm (k + 1))− cos (2wmk) + i [sin (2wm (k + 1))− sin (2wmk)]8 S.S. DRAGOMIR= −2 sin[2wm (k + 1) + 2wmk2]sin[2wm (k + 1)− 2wmk2]+i2 cos[2wm (k + 1) + 2wmk2]sin[2wm (k + 1)− 2wmk2]= −2 sin ((2k + 1)wm) sin (wm) + 2i cos ((2k + 1)wm) sin (wm)= 2i sin (wm) [cos [(2k + 1)mw] + i sin [(2k + 1)mw]]= 2i sin (wm) exp [(2k + 1)mwi] ,and then|exp (2wim (k + 1))− exp (2wimk)| = 2 |sin (wm)|Then, by (3.3) , we deduce the desired inequality (3.2) .The following particular case can be useful in practice.Corollary 2. Under the above assumptions, we have∥∥∥∥∥Fw (x¯) (m)− sin (wmn)sin (wm) exp [(n+ 1)mi]× 1nn∑k=1xk∥∥∥∥∥(3.5)≤ (n+ 1) (n− 1)323|sin (wn)|(n−1∑k=1‖∆xk‖2) 12,for all m ∈ {1, ..., n} , w 6= knpi, k ∈ Z.4. Applications for the Discrete Mellin TransformLet (X, ‖·‖) be a normed linear space over K, K = C,R, and x¯ = (x1, ..., xn) bea sequence of vectors in X.Define the Mellin transform(4.1) M (x¯) (m) :=n∑k=1km−1xk, m = 1, ..., n;of the sequence x¯ ∈ Xn.The following approximation result for the Mellin transform (4.1) holds.Theorem 7. Let X and x¯ be as above. Then we have the inequality∥∥∥∥∥M (x¯) (m)− Sm−1 (n)× 1nn∑k=1xk∥∥∥∥∥(4.2)≤ n2 − 16(n−1∑k=1[(k + 1)m−1 − km−1]p) 1p (n−1∑k=1‖∆xk‖q) 1q,where Sm (n) , m ∈ R, n ∈ N is the m−powers sum of the first n natural numbers,i.e.,(4.3) Sm (n) =n∑k=1kmand p > 1, 1p +1q = 1.ANOTHER GRU¨SS TYPE INEQUALITY 9Proof. Using the inequality (3.3) , we can state that∥∥∥∥∥n∑k=1km−1xk −n∑k=1km−1 × 1nn∑k=1xk∥∥∥∥∥≤ n2 − 16(n−1∑k=1[(k + 1)m−1 − km−1]p) 1p (n−1∑k=1‖∆xk‖q) 1qand the inequality (4.2) is obtained.Consider the following particular values of the Mellin transformµ1 (x¯) :=n∑k=1kxkandµ2 (x¯) :=n∑k=1k2xk.The following corollary holds.Corollary 3. Let X and x¯ be as specified above. Then we have the inequalities:(4.4)∥∥∥∥∥µ1 (x¯)− n+ 12n∑k=1xk∥∥∥∥∥ ≤(n2 − 1) (n− 1) 1p6(n−1∑k=1‖∆xk‖q) 1qand ∥∥∥∥∥µ2 (x¯)− (n+ 1) (2n+ 1)6n∑k=1xk∥∥∥∥∥(4.5)≤ n2 − 16[n−1∑k=1(2k − 1)p] 1p(n−1∑k=1‖∆xk‖q) 1qwhere p > 1, 1p +1q = 1.Remark 3. If we assume that p = (p1, ..., pn) is a probability distribution, i.e., pk ≥0 (k = 1, ..., n) and∑nk=1 pk = 1, then, by (4.4) and (4.5) , we get the inequalities(4.6)∣∣∣∣∣n∑k=1kpk − n+ 12∣∣∣∣∣ ≤(n2 − 1) (n− 1) 1p6(n−1∑k=1|pk+1 − pk|q) 1qand ∣∣∣∣∣n∑k=1k2pk − (n+ 1) (2n+ 1)6∣∣∣∣∣(4.7)≤ n2 − 16[n−1∑k=1(2k − 1)p] 1p(n∑k=1|pk+1 − pk|q) 1q,which can be applied for the estimation of the 1 and 2-moments of a guessing map-ping as in the paper [4]. We omit the details.10 S.S. DRAGOMIR5. Applications for PolynomialsLet (X, ‖·‖) be a normed linear space over K, K = C,R, and c¯ = (c0, ..., cn) be asequence of vectors in X.Define the polynomial P : C→ X with the coefficients c¯ byP (z) = c0 + zc1 + ...+ zncn, z ∈ C, cn 6= 0.The following approximation result for the polynomial P holds.Theorem 8. Let X, c¯ and P be as above. Then we have the inequality:∥∥∥∥P (z)− zn+1 − 1z − 1 × c0 + ...+ cnn+ 1∥∥∥∥(5.1)≤ n (n+ 1)6|z − 1|[ |z|pn − 1|z|p − 1] 1p(n−1∑k=0‖∆ck‖q) 1qfor all z ∈ C, |z| 6= 1 and p > 1, 1p + 1q = 1.Proof. Using the inequality (3.3) , we can state that∥∥∥∥∥n∑k=0zkck −n∑k=0zk × 1n+ 1n∑k=0ck∥∥∥∥∥(5.2)≤ n (n+ 1)6(n−1∑k=0∣∣zk+1 − zk∣∣p) 1p (n−1∑k=0‖∆ck‖q) 1q=n (n+ 1)6(n−1∑k=0|z|pk |z − 1|p) 1p(n−1∑k=0‖∆ck‖q) 1q=n (n+ 1)6|z − 1|(n−1∑k=0|z|pk) 1p(n−1∑k=0‖∆ck‖q) 1q=n (n+ 1)6|z − 1|( |z|pn − 1|z|p − 1) 1p(n−1∑k=0‖∆ck‖q) 1qand, asn∑k=0zk =zn+1 − 1z − 1 , z 6= 1,we obtain the desired result (5.1).The following result for the complex roots of the unity also holds:Theorem 9. Let zk := cos(kpin+1)+ i sin(kpin+1), k ∈ {0, ..., n} be the complex(n+ 1)−roots of the unity. Then we have the inequality(5.3) ‖P (zk)‖ ≤ n1+ 1p (n+ 1)3sin[kpi2 (n+ 1)](n−1∑k=0‖∆ck‖q) 1qfor all k ∈ {1, ..., n} .ANOTHER GRU¨SS TYPE INEQUALITY 11Proof. We have, from (5.2) , that∥∥∥∥∥P (zk)− zn+1k − 1zk − 1 × 1n+ 1n∑k=0ck∥∥∥∥∥(5.4)≤ n (n+ 1)6|zk − 1|(n−1∑k=1|zk|pk) 1p(n−1∑k=0‖∆ck‖q) 1q.Now, as zn+1k = 1, |zk| = 1, we get, by (5.4) , that(5.5) ‖P (zk)‖ ≤ n1+ 1p (n+ 1)6|zk − 1|(n−1∑k=0‖∆ck‖q) 1q.We observe thatzk − 1 = cos(kpin+ 1)+ i sin(kpin+ 1)− 1= −2 sin2(kpi2 (n+ 1))+ 2i sin[kpi2 (n+ 1)]cos[kpi2 (n+ 1)]= 2i sin(kpi2 (n+ 1)){cos[kpi2 (n+ 1)]+ i sin[kpi2 (n+ 1)]},and then|zk − 1| = 2∣∣∣∣sin [ kpi2 (n+ 1)]∣∣∣∣ , for all k ∈ {1, ..., n} ,and the theorem is proved.6. Applications for Lipschitzian MappingsLet (X, ‖·‖) be as above and F : X → Y a mapping defined on the normed linearspace X with values in the normed linear space Y which satisfies the Lipschitzcondition(6.1) |F (x)− F (y)| ≤ L ‖x− y‖ for all x, y ∈ X,where | · | denotes the norm of Y .The following theorem holds.Theorem 10. Let F : X → Y be as above and xi ∈ X, pi ≥ 0 (i = 1, ..., n) and∑ni=1 pi = 1. Then we have the inequality:(6.2)∣∣∣∣∣n∑i=1piF (xi)− F(n∑i=1pixi)∣∣∣∣∣ ≤ Ln∑i,j=1pipj |i− j|1q(n−1∑k=1‖∆xk‖p) 1p,where p > 1, 1p +1q = 1.Proof. As F is Lipschitzian, we have (6.1) for all x, y ∈ X. Choose x =∑ni=1 pixiand y = xj (j = 1, ..., n) to get(6.3)∣∣∣∣∣F(n∑i=1pixi)− F (xj)∣∣∣∣∣ ≤ L∥∥∥∥∥n∑i=1pixi − xj∥∥∥∥∥for all j ∈ {1, ..., n} .12 S.S. DRAGOMIRIf we multiply (6.3) by pj ≥ 0 and sum over j from 1 to n, we obtain(6.4)n∑j=1pj∣∣∣∣∣F(n∑i=1pixi)− F (xj)∣∣∣∣∣ ≤ Ln∑j=1pj∥∥∥∥∥n∑i=1pixi − xj∥∥∥∥∥ .Using the generalized triangle inequality, we have(6.5)n∑j=1pj∣∣∣∣∣F(n∑i=1pixi)− F (xj)∣∣∣∣∣ ≥∣∣∣∣∣∣F(n∑i=1pixi)−n∑j=1pjF (xj)∣∣∣∣∣∣ .By the generalized triangle inequality in normed linear space X we also haven∑j=1pj∥∥∥∥∥n∑i=1pixi − xj∥∥∥∥∥ =n∑j=1pj∥∥∥∥∥n∑i=1pi (xi − xj)∥∥∥∥∥(6.6)≤n∑i,j=1pipj ‖xi − xj‖= 2∑1≤i<j≤npipj ‖xi − xj‖ := C.As in Theorem 5, we have, for i < j that‖xi − xj‖ =∥∥∥∥∥j−1∑k=i∆xk∥∥∥∥∥ ≤j−1∑k=i‖∆xk‖and then, by Ho¨lder’s inequality for sums, we haveC ≤ 2∑1≤i<j≤npipjj−1∑k=i‖∆xk‖(6.7)≤ 2∑1≤i<j≤npipj (j − i)1q(j−1∑k=i‖∆xk‖p) 1p≤ 2(n−1∑k=1‖∆xk‖p) 1p ∑1≤i<j≤npipj (j − i)1p=(n−1∑k=1‖∆xk‖p) 1p n∑i,j=1pipj |j − i|1pand the inequality (6.2) is proved.The following corollary is a natural consequence of the above results.Corollary 4. Let X be a normed linear space and xi ∈ X (i = 1, ..., n) , pi ≥0 (i = 1, ..., n) with∑ni=1 pi = 1. Then we have(6.8) 0 ≤n∑i=1pi ‖xi‖ −∥∥∥∥∥n∑i=1pixi∥∥∥∥∥ ≤n∑i,j=1pipj |i− j|1q(n−1∑k=1‖∆xk‖p) 1p,where p > 1 and 1p +1q = 1.ANOTHER GRU¨SS TYPE INEQUALITY 13References[1] S.S. Dragomir, Gru¨ss inequality in inner product spaces, The Australian Math Soc. Gazette,26 (1999), No. 2, 66-70.[2] S.S. Dragomir, A generalization of Gru¨ss’ inequality in inner product spaces and applications,J. Math. Anal. Appl., 237 (1999), 74-82.[3] S.S. Dragomir, A Gru¨ss type integral inequality for mappings of r-Ho¨lder’s type and appli-cations for trapezoid formula, Tamkang J. of Math., 31(1) (2000), 43-47.[4] S.S. Dragomir, Some discrete inequalities of Gru¨ss type and applications in guessing theory,Honam Math. J., 21(1) (1999), 145-156.[5] S.S. Dragomir, Some integral inequalities of Gru¨ss type, Indian J. of Pure and Appl. Math.,31(4) (2000), 397-415.[6] S.S. Dragomir and G.L. Booth, On a Gru¨ss-Lupas¸ type inequality and its applications for theestimation of p-moments of guessing mappings, Mathematical Communications, 5 (2000),117-126.[7] S.S. Dragomir and I. Fedotov, An inequality of Gru¨ss’ type for Riemann-Stieltjes integral andapplications for special means, Tamkang J. of Math., 29(4)(1998), 286-292.[8] S.S. Dragomir and J.E. Pecˇaric´, On some inequalities for the moments of guessing mapping,submitted.[9] G. Gru¨ss, U¨ber das Maximum des absoluten Betrages von 1b−a∫ ba f(x)g(x)dx −1(b−a)2∫ ba f(x)dx∫ ba g(x)dx , Math. Z. , 39(1935), 215-226.[10] A.M. Fink, A treatise on Gru¨ss’ inequality, Analytic and Geometric Inequalities, 93-113,Math. Appl. 478, Kluwer Academic Publ., 1999.[11] D.S. Mitrinovic´, J.E. Pecˇaric´ and A.M. Fink, Classical and New Inequalities in Analysis,Kluwer Academic Publishers, Dordrecht, 1993.School of Communications and Informatics, Victoria University of Technology, POBox 14428, Melbourne City MC, Victoria, Australia. 8001.E-mail address: sever@matilda.vu.edu.auURL: http://rgmia.vu.edu.au/SSDragomirWeb.html

A generalization of Gr¨ uss’ inequality in inner product spaces and applications,

A Gr¨ uss type integral inequality for mappings of r-H¨ older’s type and applications for trapezoid formula,

A treatise on Gr¨ uss’ inequality, Analytic and Geometric Inequalities, 93-113,

An inequality of Gr¨ uss’ type for Riemann-Stieltjes integral and applications for special means,

Classical and New Inequalities in Analysis,

Gr¨ uss inequality in inner product spaces,

On a Gr¨ uss-Lupa¸ s type inequality and its applications for the estimation of p-moments of guessing mappings,

Peˇ cari´ c, On some inequalities for the moments of guessing mapping,

Some discrete inequalities of Gr¨ uss type and applications in guessing theory,

Some integral inequalities of Gr¨ uss type,

uss, ¨ Uber das Maximum des absoluten Betrages von 1 b−a R b a f(x)g(x)dx − (b−a)2 R b a f(x)dx R b a g(x)dx ,

Name not available

A Grüss Type Inequality for Sequences of Vectors in Normed Linear Spaces and Applications\ud

A GRU¨SS TYPE INEQUALITY FOR SEQUENCES OF VECTORSIN NORMED LINEAR SPACES AND APPLICATIONSS. S. DRAGOMIRAbstract. A discrete inequality of Gru¨ss type in normed linear spaces andapplications for the discrete Fourier transform, Mellin transform of sequences,for polynomials with coefficients in normed spaces and for vector valued Lips-chitzian mappings are given.1. IntroductionIn 1935, G. Gru¨ss [9] proved the following integral inequality which gives anapproximation of the integral of the product in terms of the product of the integralsas follows ∣∣∣∣∣ 1b− a∫ baf (x) g (x) dx− 1b− a∫ baf (x) dx · 1b− a∫ bag (x) dx∣∣∣∣∣(1.1)≤ 14(Φ− φ) (Γ− γ) ,where f , g : [a, b]→ R are integrable on [a, b] and satisfy the condition(1.2) φ ≤ f (x) ≤ Φ, γ ≤ g (x) ≤ Γfor each x ∈ [a, b] , where φ,Φ, γ,Γ are given real constants.Moreover, the constant 14 is sharp in the sense that it cannot be replaced by asmaller one.For a simple proof of (1.1) as well as for some other integral inequalities of Gru¨sstype, see Chapter X of the recent book [11] and the papers [1]-[8] and [10].In 1950, M. Biernacki, H. Pidek and C. Ryll-Nardjewski [11, Chapter X] estab-lished the following discrete version of Gru¨ss’ inequality:Theorem 1. Let a = (a1, ..., an) , b = (b1, ..., bn) be two n−tuples of real numberssuch that r ≤ ai ≤ R and s ≤ bi ≤ S for i = 1, ..., n. Then one has(1.3)∣∣∣∣∣ 1nn∑i=1aibi − 1nn∑i=1ai · 1nn∑i=1bi∣∣∣∣∣ ≤ 1n [n2 ](1− 1n[n2])(R− r) (S − s) ,where [x] denotes the integer part of x, x ∈ R.A weighted version of the discrete Gru¨ss inequality was proved by J. E. Pecˇaric´in 1979 [11, Chapter X]:Date: May, 1999.1991 Mathematics Subject Classification. Primary 26D15, 26D99; Secondary 46Bxx.Key words and phrases. Gru¨ss Inequality, Normed linear spaces, Fourier Transforms, MellinTransform, Polynomials.12 S. S. DRAGOMIRTheorem 2. Let a and b be two monotonic n−tuples and p a positive one. Then∣∣∣∣∣ 1Pnn∑i=1piaibi − 1Pnn∑i=1piai · 1Pnn∑i=1pibi∣∣∣∣∣(1.4)≤ |an − a1| |bn − b1| max1≤k≤n−1[PkP¯k+1P 2n],where Pn :=∑ni=1 pi, and P¯k+1 = Pn − Pk+1.In 1981, A. Lupas¸, [11, Chapter X] proved some similar results for the firstdifference of a as follows.Theorem 3. Let a, b be two monotonic n−tuples in the same sense and p a positiven-tuple. Thenmin1≤i≤n−1|∆ai| min1≤i≤n−1|∆bi| 1Pnn∑i=1i2pi −(1Pnn∑i=1ipi)2(1.5)≤ 1Pnn∑i=1piaibi − 1Pnn∑i=1piai · 1Pnn∑i=1pibi≤ max1≤i≤n−1|∆ai| max1≤i≤n−1|∆bi| 1Pnn∑i=1i2pi −(1Pnn∑i=1ipi)2 ,where ∆ai := ai+1 − ai is the forward first difference. If there exist the numbersa¯, a¯1, r, r1 (rr1 > 0) such that ak = a¯ + kr and bk = a¯1 + kr1, then equality holdsin (1.5).In the recent paper [6], the authors obtained the following related resultTheorem 4. Let (X, ‖·‖) be a normed linear space over K, K = R,C, xi ∈ X,αi ∈ K and pi ≥ 0 (i = 1, ..., n) such that∑ni=1 pi = 1.Then we have the inequality∥∥∥∥∥n∑i=1piαixi −n∑i=1piαin∑i=1pixi∥∥∥∥∥(1.6)≤ max1≤j≤n−1|∆αj | max1≤j≤n−1‖∆xj‖ n∑i=1i2pi −(n∑i=1ipi)2 .The inequality (1.6) is sharp in the sense that the constant c = 1 in the right handside cannot be replaced by a smaller one.In this paper we point out another inequality of Gru¨ss type and apply it inapproximating the discrete Fourier transform, the Mellin transform of sequences,for polynomials with coefficients in normed linear spaces and for vector valuedLipschitzian mappings.2. A New Discrete Inequality of Gru¨ss TypeThe following inequality of Gru¨ss type holds.A GRU¨SS TYPE INEQUALITY 3Theorem 5. Let (X, ‖·‖) be a normed linear space over K, K = R,C, xi ∈ X, αi∈ K and pi ≥ 0 (i = 1, ..., n) (n ≥ 2) such that∑ni=1 pi = 1. Then we have theinequality(2.1)∥∥∥∥∥n∑i=1piαixi −n∑i=1piαin∑i=1pixi∥∥∥∥∥ ≤ 12n∑i=1pi (1− pi)n−1∑i=1|∆αi|n−1∑i=1‖∆xi‖ ,where ∆αi := αi+1 − αi (i = 1, ..., n− 1) and ∆xi := xi+1 − xi (i = 1, ..., n− 1)are the usual forward differences.The constant 12 is sharp in the sense that it cannot be replaced by a smaller constant.Proof. Let us start with the following identity in normed linear spaces which canbe proved by direct computation [6]n∑i=1piαixi −n∑i=1piαin∑i=1pixi =12n∑i,j=1pipj (αj − αi) (xj − xi)=∑1≤i<j≤npipj (αj − αi) (xj − xi) .As i < j, we can writeαj − αi =j−1∑k=i(αk+1 − αk) =j−1∑k=i∆αkandxj − xi =j−1∑l=i(xl+1 − xl) =j−1∑l=i∆xl.Using the generalized triangle inequality, we have successively:∥∥∥∥∥n∑i=1piαixi −n∑i=1piαin∑i=1pixi∥∥∥∥∥(2.2)=∥∥∥∥∥∥∑1≤i<j≤npipjj−1∑k=i∆αkj−1∑l=i∆xl∥∥∥∥∥∥≤∑1≤i<j≤npipj∣∣∣∣∣j−1∑k=i∆αk∣∣∣∣∣∥∥∥∥∥j−1∑l=i∆xl∥∥∥∥∥≤∑1≤i<j≤npipjj−1∑k=i|∆αk|j−1∑l=i‖∆xl‖ =: A.It is obvious for all 1 ≤ i < j ≤ n− 1, we have thatj−1∑k=i|∆αk| ≤n−1∑k=1|∆αk|andj−1∑l=i‖∆xl‖ ≤n−1∑l=1‖∆xl‖4 S. S. DRAGOMIRand then(2.3) A ≤n−1∑k=1|∆αk|n−1∑l=1‖∆xl‖∑1≤i<j≤npipj .Now, let us observe that∑1≤i<j≤npipj =12 n∑i,j=1pipj −∑i=jpipj(2.4)=12 n∑i=1pin∑j=1pj −n∑i=1p2i=12n∑i=1pi (1− pi) .Using (2.2)− (2.4) , we deduce the desired inequality (2.1) .To prove the sharpness of the constant 12 , let us assume that (2.1) holds with aconstant c > 0. That is,(2.5)∥∥∥∥∥n∑i=1piαixi −n∑i=1piαin∑i=1pixi∥∥∥∥∥ ≤ cn∑i=1pi (1− pi)n−1∑i=1|∆αi|n−1∑i=1‖∆xi‖for all αi, xi, pi (i = 1, ..., n) as above and n ≥ 2.Choose in (2.1) n = 2 and compute2∑i=1piαixi −2∑i=1piαi2∑i=1pixi =122∑i,j=1pipj (αi − αj) (xi − xj)=∑1≤i<j≤2pipj (αi − αj) (xi − xj)= p1p2 (α1 − α2) (x1 − x2) .Also,2∑i=1pi (1− pi)1∑i=1|∆αi|1∑i=1‖∆xi‖ = (p1p2 + p1p2) |α1 − α2| ‖x1 − x2‖ .Substituting in (2.5) , we obtainp1p2 |α1 − α2| ‖x1 − x2‖ ≤ 2cp1p2 |α1 − α2| ‖x1 − x2‖ .If we assume that p1, p2 > 0, α1 6= α2, x1 6= x2, then we obtain c ≥ 12 , which provesthe sharpness of the constant 12 .The following corollary holds.Corollary 1. Under the above assumptions for xi, αi (i = 1, ..., n) , we have theinequality(2.6)∥∥∥∥∥ 1nn∑i=1αixi − 1nn∑i=1αi · 1nn∑i=1xi∥∥∥∥∥ ≤ 12(1− 1n) n−1∑i=1|∆αi|n−1∑i=1‖∆xi‖ ,and the constant 12 is sharp.A GRU¨SS TYPE INEQUALITY 5Considering the case of real or complex numbers is important in practical appli-cations.Corollary 2. Let αi, βi ∈ K, pi ≥ 0 (i = 1, ..., n) with∑ni=1 pi = 1. Then we havethe inequality(2.7)∣∣∣∣∣n∑i=1piαiβi −n∑i=1piαin∑i=1piβi∣∣∣∣∣ ≤ 12n∑i=1pi (1− pi)n−1∑i=1|∆αi|n−1∑i=1|∆βi| ,and the constant 12 is sharp.Remark 1. If in the above inequality we choose βi = α¯i (i = 1, ..., n) , then we get0 ≤n∑i=1pi |αi|2 −∣∣∣∣∣n∑i=1piαi∣∣∣∣∣2(2.8)≤ 12n∑i=1pi (1− pi)(n−1∑i=1|∆αi|)2,and the constant 12 is sharp.3. Applications for the Discrete Fourier TransformLet (X, ‖·‖) be a normed linear space over K, K = C,R, and x¯ = (x1, ..., xn) bea sequence of vectors in X.For a given w ∈ K, define the discrete Fourier transform(3.1) Fw (x¯) (m) :=n∑k=1xk exp (2wimk) , m = 1, ..., n.The following approximation result for the Fourier transform (3.1) holds.Theorem 6. Let (X, ‖·‖) and x¯ ∈ Xn be as above. Then we have the inequality∥∥∥∥∥Fw (x¯) (m)− sin (wmn)sin (wm) exp [(n+ 1) im]× 1nn∑k=1xk∥∥∥∥∥(3.2)≤ (n− 1)2 |sin (wm)|n−1∑i=1‖∆xi‖ ,for all m ∈ {1, ..., n} .Proof. Using the inequality (2.6) , we can state that∥∥∥∥∥n∑k=1akxk −n∑k=1ak · 1nn∑k=1xk∥∥∥∥∥(3.3)≤ n− 12n−1∑k=1|∆ak|n−1∑k=1‖∆xk‖6 S. S. DRAGOMIRfor all ak ∈ K, xk ∈ X, k = 1, ..., n.Now, choose in (3.3) , ak = exp (2wimk) to obtain∥∥∥∥∥Fw (x¯) (m)−n∑k=1exp (2wimk) · 1nn∑k=1xk∥∥∥∥∥(3.4)≤ n− 12n−1∑k=1|exp (2wim (k + 1))− exp (2wimk)|n−1∑k=1‖∆xk‖ ,for all m ∈ {1, ..., n} .However,n∑k=1exp (2wimk) = exp (2wim)×[exp (2wimn)− 1exp (2wim)− 1]= exp (2wim)×[cos (2wmn) + i sin (2wmn)− 1cos (2wm) + i sin (2wm)− 1]= exp (2wim)×[−2 sin2 (wmn) + 2i sin (wmn) cos (wmn)−2 sin2 (wm) + 2i sin (wm) cos (wm)]= exp (2wim)× sin (wmn)sin (wm)[sin (wmn)− i cos (wmn)sin (wm)− i cos (wm)]= exp (2wim)× sin (wmn)sin (wm)[cos (wmn) + i sin (wmn)cos (wm) + i sin (wm)]=sin (wmn)sin (wm)× exp (2wim)[exp (iwmn)exp (iwm)]=sin (wmn)sin (wm)× exp [2wim+ iwmn− iwm]=sin (wmn)sin (wm)× exp [(n+ 1)mi] .We observe thatexp (2wim (k + 1))− exp (2wimk)= cos (2wm (k + 1)) + i sin (2wm (k + 1))− cos (2wmk)− i sin (2wmk)= cos (2wm (k + 1))− cos (2wmk) + i [sin (2wm (k + 1))− sin (2wmk)]= −2 sin[2wm (k + 1) + 2wmk2]sin[2wm (k + 1)− 2wmk2]+i2 cos[2wm (k + 1) + 2wmk2]sin[2wm (k + 1)− 2wmk2]= −2 sin ((2k + 1)wm) sin (wm) + 2i cos ((2k + 1)wm) sin (wm)= 2i sin (wm) [cos [(2k + 1)mw] + i sin [(2k + 1)mw]]= 2i sin (wm) exp [(2k + 1)mwi] ,and then|exp (2wim (k + 1))− exp (2wimk)| = 2 |sin (wm)|A GRU¨SS TYPE INEQUALITY 7for all k = 1, ..., n− 1. Consequently,n−1∑k=1|exp (2wim (k + 1))− exp (2wimk)| = 2 |sin (wm)| (n− 1)and by (3.4) , we get the desired inequality (3.2) .4. Applications for the Discrete Mellin TransformLet (X, ‖·‖) be a normed linear space over K, K = C,R, and x¯ = (x1, ..., xn) bea sequence of vectors in X.Define the Mellin transform(4.1) M (x¯) (m) :=n∑k=1km−1xk, m = 1, ..., n;of the sequence x¯ ∈ Xn.The following result holds.Theorem 7. Let (X, ‖·‖) and x¯ ∈ Xn be as above. Then we have the inequality∥∥∥∥∥M (x¯) (m)− Sm−1 (n)× 1nn∑k=1xk∥∥∥∥∥(4.2)≤ (n− 1)(nm−1 − 1)2n−1∑k=1‖∆xk‖ ,where Sp (n) , p ∈ R, n ∈ N is the p−powers sum of the first n natural numbers,i.e.,Sp (n) =n∑k=1kp.Proof. Using the inequality (3.3) , we can state that∥∥∥∥∥n∑k=1km−1xk −n∑k=1km−1 · 1nn∑k=1xk∥∥∥∥∥≤ n− 12n−1∑k=1∣∣∣(k + 1)m−1 − km−1∣∣∣ n−1∑k=1‖∆xk‖=(n− 1) (nm−1 − 1)2n−1∑k=1‖∆xk‖ ,and the inequality (4.2) is proved.Consider the following particular values of the Mellin transformµ1 (x¯) :=n∑k=1kxkandµ2 (x¯) :=n∑k=1k2xk.The following corollary holds.8 S. S. DRAGOMIRCorollary 3. Let X and x¯ be as specified above. Then we have the inequalities:(4.3)∥∥∥∥∥µ1 (x¯)− n+ 12n∑k=1xk∥∥∥∥∥ ≤ (n− 1)22n−1∑k=1‖∆xk‖and(4.4)∥∥∥∥∥µ2 (x¯)− (n+ 1) (2n+ 1)6n∑k=1xk∥∥∥∥∥ ≤ (n− 1)2 (n+ 1)2n−1∑k=1‖∆xk‖ .Remark 2. If we assume that p = (p1, ..., pn) is a probability distribution, i.e., pk ≥0 (k = 1, ..., n) and∑nk=1 pk = 1, then, by (4.3) and (4.4) , we get the inequalities(4.5)∣∣∣∣∣n∑k=1kpk − n+ 12∣∣∣∣∣ ≤ (n− 1)22n−1∑k=1|pk+1 − pk|and(4.6)∣∣∣∣∣n∑k=1k2pk − (n+ 1) (2n+ 1)6∣∣∣∣∣ ≤ (n− 1)2 (n+ 1)2n−1∑k=1|pk+1 − pk| ,which have been obtained in [4] and applied for the estimation of the 1 and 2-moments of a guessing mapping.5. Applications for PolynomialsLet (X, ‖·‖) be a normed linear space over K, K = C,R, and c¯ = (c0, ..., cn) be asequence of vectors in X.Define the polynomial P : C→ X with the coefficients c¯ byP (z) = c0 + zc1 + ...+ zncn, z ∈ C, cn 6= 0.The following approximation result for the polynomial P holds.Theorem 8. Let X, c¯ and P be as above. Then we have the inequality:(5.1)∥∥∥∥P (z)− zn+1 − 1z − 1 × c0 + ...+ cnn+ 1∥∥∥∥ ≤ n |z − 1|2 (|z| − 1) (|z|n − 1)n−1∑k=0‖∆ck‖for all z ∈ C, |z| 6= 1.Proof. Using the inequality (3.3) , we can state that∥∥∥∥∥n∑k=0zkck −n∑k=0zk × 1n+ 1n∑k=0ck∥∥∥∥∥(5.2)≤ n2n−1∑k=0∣∣zk+1 − zk∣∣ n−1∑k=0‖∆ck‖=n2n−1∑k=0|z|k |z − 1|n−1∑k=0‖∆ck‖=n |z − 1|2 (|z| − 1) (|z|n − 1)n−1∑k=0‖∆ck‖ ,and, as∑nk=0 zk = zn+1−1z−1 , z 6= 1, the inequality (5.1) is proved.A GRU¨SS TYPE INEQUALITY 9The following result for the complex roots of the unity also holds:Theorem 9. Let zk := cos(kpin+1)+ i sin(kpin+1), k ∈ {0, ..., n} be the complex(n+ 1)−roots of the unity. Then we have the inequality(5.3) ‖P (zk)‖ ≤ n2 sin[kpi2 (n+ 1)] n−1∑k=0‖∆ck‖for all k ∈ {1, ..., n} .Proof. As in the proof of Theorem 8, we have∥∥∥∥∥P (z)− zn+1 − 1z − 1 × 1n+ 1n∑k=0ck∥∥∥∥∥(5.4)≤ n2n−1∑m=0|z|m |z − 1|n−1∑k=0‖∆ck‖ .If we choose z = zk, k ∈ {1, ..., n} , we have |zk|m = 1, zn+1k = 1 and then, by (5.4)we deduce(5.5) ‖P (zk)‖ ≤ n22|zk − 1|n−1∑k=0‖∆ck‖ .However,zk − 1 = cos(kpin+ 1)+ i sin(kpin+ 1)− 1= −2 sin2(kpi2 (n+ 1))+ 2i sin[kpi2 (n+ 1)]cos[kpi2 (n+ 1)]= 2i sin(kpi2 (n+ 1))[cos[kpi2 (n+ 1)]+ i sin[kpi2 (n+ 1)]]and then|zk − 1| = 2 sin[kpi2 (n+ 1)], for all k ∈ {1, ..., n} .Using (5.5) , we deduce the desired inequality (5.3) .Corollary 4. Let P (z) =∑nk=0 ckzk be a polynomial with real coefficients satisfy-ing the condition c0 ≤ c1 ≤ ... ≤ cn. Then we have the inequality(5.6) |P (zk)| ≤ n2 sin[kpi2 (n+ 1)](cn − c0) ,for all k ∈ {1, ..., n} , where zk are as in Theorem 9.6. Applications for Lipschitzian MappingsLet (X, ‖·‖) be as above and F : X → Y a mapping defined on the normed linearspace X with values in the normed linear space Y which satisfies the Lipschitziancondition(6.1) |F (x)− F (y)| ≤ L ‖x− y‖ for all x, y ∈ X,where | · | denotes the norm on Y .The following theorem holds.10 S. S. DRAGOMIRTheorem 10. Let F : X → Y be as above and xi ∈ X, pi ≥ 0 (i = 1, ..., n) and∑ni=1 pi = 1. Then we have the inequality:(6.2)∣∣∣∣∣n∑i=1piF (xi)− F(n∑i=1pixi)∣∣∣∣∣ ≤ Ln∑i=1pi (1− pi)n−1∑k=1‖∆xk‖ .Proof. As F is Lipschitzian, we have (6.1) for all x, y ∈ X. Choose x =∑ni=1 pixiand y = xj (j = 1, ..., n) to get(6.3)∣∣∣∣∣F(n∑i=1pixi)− F (xj)∣∣∣∣∣ ≤ L∥∥∥∥∥n∑i=1pixi − xj∥∥∥∥∥for all j ∈ {1, ..., n} .If we multiply (6.3) by pj ≥ 0 and sum over j from 1 to n, we obtain(6.4)n∑j=1pj∣∣∣∣∣F(n∑i=1pixi)− F (xj)∣∣∣∣∣ ≤ Ln∑j=1pj∥∥∥∥∥n∑i=1pixi − xj∥∥∥∥∥ .Using the generalized triangle inequality, we have(6.5)n∑j=1pj∣∣∣∣∣F(n∑i=1pixi)− F (xj)∣∣∣∣∣ ≥∣∣∣∣∣∣F(n∑i=1pixi)−n∑j=1pjF (xj)∣∣∣∣∣∣ .By the generalized triangle inequality in the normed space X, we also haven∑j=1pj∥∥∥∥∥n∑i=1pixi − xj∥∥∥∥∥ =n∑j=1pj∥∥∥∥∥n∑i=1pi (xi − xj)∥∥∥∥∥(6.6)≤n∑i,j=1pipj ‖xi − xj‖= 2∑1≤i<j≤npipj ‖xi − xj‖ := B.As in the proof of Theorem 3, we have, for i < j‖xi − xj‖ =∥∥∥∥∥j−1∑k=i∆xk∥∥∥∥∥ ≤j−1∑k=i‖∆xk‖and thenB ≤ 2∑1≤i<j≤npipjj−1∑k=i‖∆xk‖≤ 2∑1≤i<j≤npipjn−1∑k=1‖∆xk‖ .Since ∑1≤i<j≤npipj =12n∑i=1pi (1− pi)then we get, by (6.4)− (6.6) , the desired inequality (6.2) .The following corollary is a natural consequence of the above results.A GRU¨SS TYPE INEQUALITY 11Corollary 5. Let xi ∈ X and pi be as above. Then we have the inequality:0 ≤n∑i=1pi ‖xi‖ −∥∥∥∥∥n∑i=1pixi∥∥∥∥∥ ≤n∑i=1pi (1− pi)n−1∑k=1‖∆xk‖ .References[1] S.S. Dragomir, Gru¨ss inequality in inner product spaces, The Australian Math Soc. Gazette,26 (1999), No. 2, 66-70.[2] S.S. Dragomir, A generalization of Gru¨ss’ inequality in inner product spaces and applications,J. Math. Anal. Appl., 237 (1999), 74-82.[3] S.S. Dragomir, A Gru¨ss type integral inequality for mappings of r-Ho¨lder’s type and appli-cations for trapezoid formula, Tamkang J. of Math., 31(1) (2000), 43-47.[4] S.S. Dragomir, Some discrete inequalities of Gru¨ss type and applications in guessing theory,Honam Math. J., 21(1) (1999), 145-156.[5] S.S. Dragomir, Some integral inequalities of Gru¨ss type, Indian J. of Pure and Appl. Math.,31(4) (2000), 397-415.[6] S.S. Dragomir and G.L. Booth, On a Gru¨ss-Lupas¸ type inequality and its applications for theestimation of p-moments of guessing mappings, Mathematical Communications, 5 (2000),117-126.[7] S.S. Dragomir and I. Fedotov, An inequality of Gru¨ss’ type for Riemann-Stieltjes integral andapplications for special means, Tamkang J. of Math., 29(4)(1998), 286-292.[8] S.S. Dragomir and J.E. Pecˇaric´, On some inequalities for the moments of guessing mapping,submitted.[9] G. Gru¨ss, U¨ber das Maximum des absoluten Betrages von 1b−a∫ ba f(x)g(x)dx −1(b−a)2∫ ba f(x)dx∫ ba g(x)dx , Math. Z. , 39(1935), 215-226.[10] A.M. Fink, A treatise on Gru¨ss’ inequality, Analytic and Geometric Inequalities, 93-113,Math. Appl. 478, Kluwer Academic Publ., 1999.[11] D.S. Mitrinovic´, J.E. Pecˇaric´ and A.M. Fink, Classical and New Inequalities in Analysis,Kluwer Academic Publishers, Dordrecht, 1993.School of Communications and Informatics, Victoria University of Technology, POBox 14428, Melbourne City, MC 8001 AustraliaE-mail address: sever@matilda.vu.edu.auURL: http://rgmia.vu.edu.au/SSDragomirWeb.html

Another Grüss Type Inequality for Sequences of Vectors in Normed Linear Spaces and Applications

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