In this paper we point out a converse result of the celebrated Jensen inequality for differentiable convex mappings of several variables and apply it to counterpart well-known analytic inequalities. Applications to Shannon's

and Rényi's entropy mappings are also given

Dragomir, Sever S

English

Victoria University Institutional Repository (VUIR)

A Converse of the Jensen Inequality for Convex Mappings of Several Variables and Applications

Victoria University Eprints Repository

A CONVERSE OF THE JENSEN INEQUALITY FOR CONVEXMAPPINGS OF SEVERAL VARIABLES AND APPLICATIONSS. S. DRAGOMIRAbstract. In this paper we point out a converse result of the celebratedJensen inequality for differentiable convex mappings of several variables andapply it to counterpart well-known analytic inequalities. Applications to Shan-non’s and Re´nyi’s entropy mappings are also given.1. IntroductionLet f : X → R be a convex mapping defined on the linear space X and xi ∈ X,pi ≥ 0 (i = 1, ...,m) with Pm :=∑mi=1 pi > 0.The following inequality is well known in the literature as Jensen’s inequalityf(1Pmm∑i=1pixi)≤ 1Pmm∑i=1pif (xi) .(1.1)There are many well known inequalities which are particular cases of Jensen’s in-equality such as the weighted arithmetic mean-geometric mean- harmonic meaninequality, the Ky Fan inequality, the Ho¨lder inequality, etc. For a comprehensivelist of recent results on the Jensen inequality, see the book [1] and the papers [2]-[14]where further results are givenIn this paper, we point out a converse inequality for Jensen’s result (1.1) forthe case of differentiable convex mappings whose partial derivatives are bounded.Applications for some particular inequalities and for the Shannon and Re´nyi entropymappings are also given.2. A Converse InequalityThe following converse of Jensen’s inequality holds.Theorem 1. Let f : Rn → R be a differentiable convex mapping and xi ∈ Rn,i = 1, ...,m. Suppose that there exist the vectors ψ, φ ∈ Rn such that ψ ≤ xi ≤ φ (theorder is considered on the co-ordinates) and m,M ∈ Rn are such that m ≤ ∂f(xi)∂x ≤M for all i ∈ {1, ...,m} . Then for all pi ≥ 0 (i = 1, ...,m) with Pm :=∑mi=1 pi > 0,we have the inequality0 ≤ 1Pmm∑i=1pif (xi)− f(1Pmm∑i=1pixi)≤ 14‖Φ− φ‖ ‖M −m‖ ,(2.1)where ‖·‖ is the usual Euclidean norm on Rn.Date: 16th July, 1999.1991 Mathematics Subject Classification. Primary: 26D15; Secondary: 94Xxx.Key words and phrases. Convex functions, Jensen’s inequality, Entropy mapping.12 S. S. DRAGOMIRProof. Firstly, and for the sake of completeness, we prove the following inequalityfor convex functions obtained by Dragomir and Goh in [14]:0 ≤ 1Pmm∑i=1pif (xi)− f(1Pmm∑i=1pixi)(2.2)≤ 1Pmm∑i=1pi 〈xi,∇f (xi)〉 −〈1Pmm∑i=1pixi,1Pmm∑i=1pi∇f (xi)〉,where 〈·, ·〉 is the usual inner product on Rn and∇f (x) = ∂f (x)∂x:=(∂f (x)∂x1, ...,∂f (x)∂xn)is the vector of the partial derivatives, x =(x1, ..., xn) ∈ Rn.As f : Rn → R is differentiable convex, we have the inequalityf (x)− f (y) ≥ 〈∇f (y) , x− y〉 , for all x, y ∈ Rn.(2.3)Choose in (2.3), x = 1Pm∑mi=1 pixi and y = xj to obtainf(1Pmm∑i=1pixi)− f (xj) ≥〈∇f (xj) , 1Pmm∑i=1pixi − xj〉(2.4)for all j ∈ {1, ..., n} .If we multiply (2.4) by pj ≥ 0 and sum over j from 1 to m, we obtainPmf(1Pmm∑i=1pixi)−m∑j=1pjf (xj)≥ 1Pm〈m∑j=1pj∇f (xj) ,m∑i=1pixi〉−m∑j=1〈∇f (xj) , xj〉 .Dividing by Pm > 0, we obtain (2.2).A simple calculation shows that1Pmm∑i=1pi 〈xi,∇f (xi)〉 −〈1Pmm∑i=1pixi,1Pmm∑i=1pi∇f (xi)〉(2.5)=12P 2mm∑i,j=1pipj 〈xi − xj ,∇f (xi)−∇f (xj)〉 .Taking the modulus in both parts of (2.5), and accounting for the fact that theleft hand side is positive (by (2.2)), we obtain, by Schwartz’s inequality in innerproduct spaces, i.e., we recall it |〈a, b〉| ≤ ‖a‖ ‖b‖ , a, b ∈ Rn, that1Pmm∑i=1pi 〈xi,∇f (xi)〉 −〈1Pmm∑i=1pixi,1Pmm∑i=1pi∇f (xi)〉(2.6)≤ 12P 2mm∑i,j=1pipj |〈xi − xj ,∇f (xi)−∇f (xj)〉|≤ 12P 2mm∑i,j=1pipj ‖xi − xj‖ ‖∇f (xi)−∇f (xj)‖ .3Using the Cauchy-Buniakowsky-Schwartz inequality for double sums, we can statethat12P 2mm∑i,j=1pipj ‖xi − xj‖ ‖∇f (xi)−∇f (xj)‖(2.7)≤ 12P 2mm∑i,j=1pipj ‖xi − xj‖2 12 × 12P 2mm∑i,j=1pipj ‖∇f (xi)−∇f (xj)‖2 12 .As a simple calculation shows that12P 2mm∑i,j=1pipj ‖xi − xj‖2 = 1Pmm∑i=1pi ‖xi‖2 −∥∥∥∥∥ 1Pmm∑i=1pixi∥∥∥∥∥2and12P 2mm∑i,j=1pipj ‖∇f (xi)−∇f (xj)‖2=1Pmm∑i=1pi ‖∇f (xi)‖2 −∥∥∥∥∥ 1Pmm∑i=1∇f (xi)∥∥∥∥∥2,we can state, by (2.6) and (2.7), that1Pmm∑i=1pi 〈xi,∇f (xi)〉 −〈1Pmm∑i=1pixi,1Pmm∑i=1pi∇f (xi)〉(2.8)≤ 1Pmm∑i=1pi ‖xi‖2 −∥∥∥∥∥ 1Pmm∑i=1pixi∥∥∥∥∥2 12× 1Pmm∑i=1pi ‖∇f (xi)‖2 −∥∥∥∥∥ 1Pmm∑i=1∇f (xi)∥∥∥∥∥2 12 .Now, let us observe that, by a simple calculation1Pmm∑i=1pi ‖xi‖2 −∥∥∥∥∥ 1Pmm∑i=1pixi∥∥∥∥∥2(2.9)=〈φ− 1Pmm∑i=1pixi,1Pmm∑i=1pixi − ψ〉− 1Pmm∑i=1pi 〈φ− xi, xi − ψ〉 .As ψ ≤ xi ≤ φ (i ∈ {1, ...,m}) , then 〈φ− xi, xi − ψ〉 ≥ 0 for all i ∈ {1, ...,m} andthenm∑i=1pi 〈φ− xi, xi − ψ〉 ≥ 04 S. S. DRAGOMIRand, by (2.9), we obtain1Pmm∑i=1pi ‖xi‖2 −∥∥∥∥∥ 1Pmm∑i=1pixi∥∥∥∥∥2(2.10)≤〈φ− 1Pmm∑i=1pixi,1Pmm∑i=1pixi − ψ〉.It is known that if y, z ∈ Rn, then4 〈z, y〉 ≤ ‖z + y‖2(2.11)with equality iff z = y.Now, if we apply (2.11) for the vectors z = φ− 1PM∑mi=1 pixi, y =1Pm∑mi=1 pixi−ψ, we deduce 〈φ− 1Pmm∑i=1pixi,1Pmm∑i=1pixi − ψ〉≤ 14‖φ− ψ‖2and then, by (2.9) - (2.10), we deduce that1Pmm∑i=1pi ‖xi‖2 −∥∥∥∥∥ 1Pmm∑i=1pixi∥∥∥∥∥2≤ 14‖φ− ψ‖2 .(2.12)Similarly, we can state that1Pmm∑i=1pi ‖∇f (xi)‖2 −∥∥∥∥∥ 1Pmm∑i=1∇f (xi)∥∥∥∥∥2≤ 14‖M −m‖2 .(2.13)Finally, by (2.8) and (2.12) - (2.13), we deduce1Pmm∑i=1pi 〈xi,∇f (xi)〉 −〈1Pmm∑i=1pixi,1Pmm∑i=1pi∇f (xi)〉(2.14)≤ 14‖Φ− φ‖ ‖M −m‖ ,which, by (2.2), gives the desired inequality (2.1).Remark 1. A similar result for integrals can be stated, but we omit the details.Remark 2. The conditionsψ ≤ xi ≤ φ, m ≤ ∂f (xi)∂x≤M (i = 1, ...,m)(2.15)can be replaced by the more general conditionsm∑i=1pi 〈φ− xi, xi − ψ〉 ≥ 0 andm∑i=1pi〈M − ∂f (xi)∂x,∂f (xi)∂x−m〉≥ 0(2.16)and the conclusion (2.1) will still be valid.Remark 3. Even if the new inequality (2.1) is not as sharp as the Dragomir-Gohinequality (2.2), it may be more useful in practice when only some bounds of the5partial derivatives ∂f∂x and of the vectors xi (i = 1, ...,m) are known. Namely, itprovides the opportunity to estimate the difference1Pmm∑i=1pif (xi)− f(1Pmm∑i=1pixi)=: ∆ (f, x, p)when the quantities ‖φ− ψ‖ and ‖M −m‖ are known.For example, if the partial derivatives ∂f∂x are bounded, i.e., there exists m,M ∈Rn such that m ≤ ∂f∂x ≤ M on the co-ordinates, and if we choose the vector xi(i = 1, ...,m) not “very far” from a constant vector x0, i.e., ‖φ− ψ‖ ≤ 4ε‖M−m‖ ,ε > 0, then, by (2.1), we can conclude that0 ≤ ∆ (f, x, p) ≤ ε.The case of convex mappings of a real variable can be stated as follows [20].Corollary 1. Let f : R→ R be a differentiable convex mapping and xi ∈ I for alli ∈ {1, ...,m} . Then we have the inequality:0 ≤ 1Pmm∑i=1pif (xi)− f(1Pmm∑i=1pixi)(2.17)≤ 14(M −m) (f ′ (M)− f ′ (m)) ,where pi > 0 (i = 1, ...,m) and Pm :=∑mi=1 pi > 0.The proof is obvious by the above findings, taking into account that the mappingf ′ is monotonic nondecreasing, and then f ′ (m) ≤ f ′ (xi) ≤ f ′ (M) for all i ∈{1, ...,m} .3. Applications for Weighted MeansConsider the classical weighted means:An (p¯, x) : =n∑i=1pixi - the arithmetic mean,Gn (p¯, x) : =n∏i=1xpii - the geometric mean,Hn (p¯, x) : =1∑ni=1pixi- the harmonic mean,provided that xi > 0(i = 1, n)and pi (i = 1, ..., n) is a probability distribution,i.e.,∑ni=1 pi = 1.The following inequality is well-known in the literature as the arithmetic mean-geometric mean-harmonic mean inequalityAn (p¯, x) ≥ Gn (p¯, x) ≥ Hn (p¯, x) ,(3.1)with equality iff x1 = ... = xn (for pi > 0, i = 1, ..., n).The following corollaries hold.6 S. S. DRAGOMIRCorollary 2. Let 0 < m ≤ xi ≤M <∞, pi > 0 (i = 1, ..., n) . Then1 ≤ An (p¯, x)Gn (p¯, x)≤ exp[(M −m)24mM].(3.2)Equality holds in (3.2) simultaneously iff x1 = ... = xn.The proof follows by the inequality (2.17) for f (x) = − lnx, x > 0.Corollary 3. Let 0 < m ≤ yi ≤M <∞, pi > 0 (i = 1, ..., n) . Then1 ≤ Gn (p¯, y¯)Hn (p¯, y¯)≤ exp[(M −m)24mM].(3.3)Equality holds iff y1 = ... = yn.Corollary 4. Let p ≥ 1 and 0 ≤ m ≤ xi ≤M <∞, pi > 0 (i = 1, .., n) . Then0 ≤n∑i=1pixpi −(n∑i=1pixi)p≤ p4(M −m) (Mp−1 −mp−1) .(3.4)Equality holds iff x1 = ... = xn.The proof follows by (2.17) for the mapping f (x) = xp, p ≥ 1, x ≥ 0.Finally, we haveCorollary 5. Let pi, xi be as in Corollary 3. Then1 ≤∏ni=1 xpixii[An (p¯, x)]An(p¯,x)≤(Mm) 14 (M−m).(3.5)Equality holds iff x1 = ... = xn.The proof follows by (2.1) for the mapping f (x) = x lnx, x > 0.4. Applications for Shannon’s EntropyLet X be a random variable with the range R = {x1, ..., xn} and the probabilitydistribution p1, ..., pn (pi > 0, i = 1, ..., n) . Define the Shannon entropy mappingH (X) := −n∑i=1pi ln pi.The following theorem is well known in the literature and concerns the maximumpossible value of H (X) in terms of the size of R [15, p. 27].Theorem 2. Let X be defined as above. Then0 ≤ H (X) ≤ lnn.(4.1)Furthermore, H (X) = 0 iff pi = 1 for some i and H (X) = lnn iff pi = 1n for alli ∈ {1, ..., n} .In the recent paper [14], Dragomir and Goh proved the following counterpartresult.Theorem 3. Let X be defined as above. Then0 ≤ lnn−H (X) ≤∑1≤i<j≤n(pi − pj)2 .(4.2)Equality holds simultaneously in both inequalities iff pi = 1n for all i ∈ {1, ..., n} .7We give here the following lemma which concerns an analytic inequality for thelog−map with applications for entropy.Lemma 1. Let 0 < m ≤ ξi ≤ M < ∞, pi > 0 (i = 1, ..., n) with∑ni=1 pi = 1.Then0 ≤ ln(n∑i=1piξi)−n∑i=1pi ln ξi ≤(M −m)24mM.(4.3)The proof follows by Corollary 1, choosing f (x) = − lnx and xi = ξi, i = 1, ..., n.The above lemma provides the following converse inequality for the Shannonentropy mapping (see also [17, p. ]):Theorem 4. Let X be as above and put p := mini=1,n pi and P := maxi=1,n pi.Then we have0 ≤ lnn−H (X) ≤ (P − p)24pP.(4.4)Proof. Choose in the above lemma ξi =1pi∈[1P ,1p]and m = 1P ,M =1p to get thedesired inequality (4.4).Another analytic inequality which can be applied for the entropy mapping isembodied in the following lemma.Lemma 2. Let 0 < m ≤ ξi ≤ M < ∞, pi > 0 (i = 1, ..., n) with∑ni=1 pi = 1.Then0 ≤n∑i=1piξi ln ξi −n∑i=1piξi ln(n∑i=1piξi)(4.5)≤ 14(M −m) (lnM − lnm)≤ 14· (M −m)2√mM.Proof. The first inequality follows by Corollary 1, choosing f (x) = x lnx, which isa convex mapping on (0,∞) , and xi = ξi, i = 1, ..., n.The second inequality follows by the celebrated inequality between the geometricmean G (a, b) :=√ab and the logarithmic meanL (a, b) :=a if b = ab−aln b−ln a if b 6= a, a, b > 0which states thatG (a, b) ≤ L (a, b) , a, b > 0i.e.,ln b− ln ab− a ≤1√ab.8 S. S. DRAGOMIRChoosing b = M, a = m, we obtain14(M −m) (lnM − lnm) ≤ 14(M −m) (lnM − lnm)√Mm=14√Mm(M −m)2 .The lemma provides the following converse inequality for the entropy mapping[18].Theorem 5. Let X be as in Theorem 4. Then we have0 ≤ lnn−H (X) ≤ n4(P − p) (lnP − ln p) ≤ n4· (P − p)2√pP.(4.6)Proof. Firstly, let us chose pi = 1n in (4.5) to get0 ≤ 1nn∑i=1ξi ln ξi −1nn∑i=1ξi ln(1nn∑i=1ξi)(4.7)≤ 14(M −m) (lnM − lnm)≤ 14√Mm(M −m)2 .Now, if in (4.7) we assume that ξi = pi ∈ [p, P ] , then we obtain0 ≤ 1nlnn− 1nH (X) ≤ 14(P − p) (lnP − ln p) ≤ 14· (P − p)2√pP,from where results (4.6).5. Applications for Re´nyi’s entropyThe Re´nyi entropy of order α, α ∈ (0, 1) ∪ (1,∞) is defined by [19]Hα (X) :=11− α ln(n∑i=1pαi).(5.1)Using Jensen’s inequality for convex mappings applied for f (x) = − ln (x) , we haveln(n∑i=1pixi)≥n∑i=1pi lnxi, xi, pi > 0 (i = 1, ..., n) ,n∑i=1pi = 1.(5.2)If we choose xi := pα−1i (i = 1, ..., n) in (5.2), we obtainln(n∑i=1pαi)≥ (α− 1)n∑i=1pi ln pi,which is equivalent to(1− α) [Hα (X)−H (X)] ≥ 0.(5.3)Now, if α ∈ (0, 1) , then Hα (X) ≤ H (X) and if α > 1, then Hα (X) ≥ H (X) .Equality holds in (5.3) iff (pi)i=1,n is a uniform distribution and this fact followsby the strict convexity of − ln (·) .We can now point out a counterpart result for (5.3).9Theorem 6. With the above assumptions, we have(1− α) [Hα (X)−H (X)] ≤ 14(Pα−1 − pα−1)2pα−1Pα−1,(5.4)provided that 0 < p ≤ pi ≤ P < 1 (i = 1, ..., n) .Proof. We use Lemma 1 for ξi := pα−1i and take into account that, for α ∈ (0, 1) ,we have Pα−1 ≤ ξi ≤ pα−1, (i = 1, ..., n) and, for α ∈ (1,∞) , we have pα−1 ≤ ξi ≤Pα−1 (i = 1, ..., n) . Choosing in the first case m = Pα−1, M = pα−1 and in thesecond case m = pα−1, M = Pα−1, we obtain the same upper bound(M −m)2mM=(Pα−1 − pα−1)2pα−1Pα−1.Now, let us remark that a particular case of (4.3) for pi = 1n (i = 1, ..., n) statesthat0 ≤ ln(1nn∑i=1ξi)− 1nn∑i=1ln (ξi) ≤(M −m)24mM,(5.5)provided that 0 < m ≤ ξi ≤M <∞ (i = 1, ..., n) .This inequality allows us to prove the following inequality for the Re´nyi entropy:Theorem 7. With the assumptions of Theorem 6, we have the inequality0 ≤ (1− α)Hα (X)− lnn− α lnGn (p)(5.6)≤ n4· (Pα − pα)2pαPα,where Gn (p) is the geometric mean of pi (i = 1, ..., n) , i.e., we recall that Gn (p) =(∏ni=1 pi)1n .Proof. We choose in the inequality (5.8) ξi = npαi (i = 1, ..., n) and observe thatnpα ≤ ξi ≤ nPα, so we can have m = npα and M = nPα in (5.5), getting thedesired inequality.If we assume that α ∈ (0, 1) and apply Corollary 1 for the convex mappingf (x) = −xα, we deduce the following inequality0 ≤(n∑i=1pixi)α−n∑i=1pixαi ≤α4(M −m) (mα−1 −Mα−1) ,(5.7)provided that 0 < m ≤ xi ≤M <∞, pi > 0 (i = 1, ..., n) and∑ni=1 pi = 1.If in (5.7) we put pi = 1n (i = 1, ..., n) , then we obtain0 ≤ 1nα(n∑i=1xi)α− 1nn∑i=1xαi ≤α4(M −m) (mα−1 −Mα−1) .(5.8)The following inequality for the α-Re´nyi entropy holds.Theorem 8. If 0 < p ≤ pi ≤ P < 1 (i = 1, ..., n) and α ∈ (0, 1) , then we have theinequality0 ≤ n1−α − exp [(1− α)Hα (X)] ≤ α4 n (P − p)(pα−1 − Pα−1) .(5.9)10 S. S. DRAGOMIRProof. Choosing xi = pi (i = 1, ..., n) in (5.8), we deduce0 ≤ 1nα− 1nn∑i=1pαi ≤α4(P − p) (pα−1 − Pα−1) .(5.10)Taking into account that∑ni=1 pαi = exp [(1− α)Hα (X)] , then from (5.10) wededuce the desired inequality (5.9).Now, if we define by E (X) :=∑ni=1 p2i , the informational energy of the randomvariable X, then we may also point out the following inequality.Theorem 9. If pi and α are as in Theorem 8, then we have the inequality0 ≤ Eα (X)− exp [−αHα+1 (X)] ≤ α4 (P − p)(pα−1 − Pα−1) .(5.11)The proof follows by the inequality (5.7), choosing xi = pi, i = 1, ..., n, and weomit the details.Now assume that α ∈ (1,∞) . Then, applying Corollary 1 for the convex mappingf (x) = xα, we deduce the following inequality0 ≤n∑i=1pixαi −(n∑i=1pixi)α≤ α4(M −m) (mα−1 −Mα−1) ,(5.12)provided that 0 < m ≤ xi ≤M <∞, pi > 0 (i = 1, ..., n) and∑ni=1 pi = 1.Finally, using a similar argument to the one above, we can state the followingtheorem.Theorem 10. Let α ∈ (1,∞) and 0 < p ≤ pi ≤ P < 1 (i = 1, ..., n) . Then wehave the inequalities0 ≤ exp [−αHα (X)]− Eα (X) ≤ α4 (P − p)(Pα−1 − pα−1)and0 ≤ exp [(1− α)Hα (X)]− n1−α ≤ α4 n (P − p)(Pα−1 − pα−1) .References[1] J. E. PECˇARIC´, F. PROSCHAN and Y.L. TONG, Convex Functions, Partial Orderings andStatistical Applications, Academic Press, 1992.[2] S.S. DRAGOMIR, An improvement of Jensen’s inequality, Bull. Math. Soc. Sci. Math. Ro-mania., 34 (1990), 291 - 296.[3] S.S. DRAGOMIR, Some refinements of Jensen’s inequality, J. Math. Anal. Appl., 168 (1992),518 -522.[4] S.S. DRAGOMIR, Some refinements of Ky Fan’s inequality, J. Math. Anal. Appl., 163(1992), 317 -321.[5] S.S. DRAGOMIR, On some refinements of Jensen’s inequality and applications, UtilitasMath., 43 (1993), 235 - 243.[6] S.S. DRAGOMIR, Two mappings in connection with Jensen’s inequality, Extracta Math., 8(2-3) (1993), 102 - 105.[7] S.S. DRAGOMIR, A further improvement of Jensen’s inequality, Tamkang J. Math., 25 (1)(1994), 29 - 36.[8] S.S. DRAGOMIR, J. E. PECˇARIC´ and L.E. PERSSON, Properties of some functionalsrelated to Jensen’s inequality, Acta Math. Hungarica, 20 (1-2 (1998), 129 - 143.[9] S.S. DRAGOMIR and C.J. GOH, Some counterpart inequalities for a functional associatedwith Jensen’s inequality, J. of Ineq. and Appli., 1 (1997), 311 - 325.[10] S.S. DRAGOMIR and S. FITZPARTICK, s - Orlicz convex functions in linear spaces andJensen’s discrete inequality, J. Math. Anal. Appl., 210 (1997), 419 - 439.11[11] S.S. DRAGOMIR, C.E.M. PEARCE and J. E. PECˇARIC´, On Jensen’s and related inequal-ities for isotonic sublinear functionals, Acta Sci. math., (Szeged), 61 (1995), 373 - 382.[12] S.S. DRAGOMIR, A new improvement of Jensen’s inequality, Indian J. Pure Appl. Math.,26 (10) (1995), 959 - 968.[13] S.S. DRAGOMIR and N.M. IONESCU, Some Converse of Jensen’s inequality and applica-tions, Anal. Num. Theor. Approx. ( Cluj-Napoca), 23 (1994), 71 - 78.[14] S.S. DRAGOMIR and C.J. GOH, A counterpart of Jensen’s discrete inequality for differen-tiable convex mappings and applications in Information Theory, Mathl. Comput. Modelling,24 (2) (1996), 1 - 11.[15] R.J. McELIECE, The Theory of Information and Coding, Addison Wesley Publishing Com-pany, Reading, 1977.[16] S.S. DRAGOMIR and C.J. GOH, Some bounds on entropy measures in information theory,Appl. Math. Lett., 10 (3) (1997), 23 - 28.[17] M. MATIC´, Jensen’s Inequality and Applications in Information Theory (in Croatian), Ph.D.Dissertation, Split, Croatia, 1998.[18] S.S. DRAGOMIR, C.E.M. PEARCE and J. E. PECˇARIC´, New inequalities for logarithmicmap and their applications for entropy and mutual information, submitted.[19] A. RE´NYI, On measures of entropy and information, Proc. Fourth Berkley Symp. Math.Statist. Prob., Vol. 1, 1961, pp. 547 - 561, Univ. of California Press, Berkley.[20] S.S. DRAGOMIR, A converse result for Jensen’s discrete inequality via Gru¨ss’ inequality andapplications in information theory, submitted.School of Communications and Informatics, Victoria University of Technology, POBox 14428, Melbourne City MC 8001, AustraliaE-mail address: sever@matilda.vu.edu.au

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In this paper we point out a converse result of the celebrated Jensen inequality for differentiable convex mappings of several variables and apply it to counterpart well-known analytic inequalities. Applications to Shannon's\ud
and Rényi's entropy mappings are also given

Name not available