 Applied Mathematics, 2011, 2, 899-903 doi:10.4236/am.2011.27120 Published Online July 2011 (http://www.SciRP.org/journal/am) Copyright © 2011 SciRes. AM Refinements to Hadamard’s Inequality for Log-Convex Functions Waadallah T. Sulaiman Department of C om put er Engineering, College of Engineering, University of Mosul, Mosul, Iraq E-mail: waadsulaiman@hotmail.com Received April 25, 2011; revised May 25, 2011; accepted May 28, 2011 Abstract In this paper we show that a log-convex function satisfies Hadamard’s inequality, as well as we give an ex- tension for this result in several directions. Keywords: Log-Convex Functions, Hadamard’s Inequality, Integral Inequality 1. Introduction Let be a convex mapping of the interval :fI of real numbers and a, bI with a < b. The fol- lowing double inequality 1d 22 b a afb ab ffxx ba (1.1) is known in the literature as Hadamard’s inequality. In [1], Fejer generalized the inequality (1.1) by proving that if :,gab is nonnegative, integrable and symmet- ric to 2 ab x, and if is convex on [a,b], then dd 2 d. 2 bb aa b a ab fgxxfxgxx fa fb xx (1.2) A positive function f is log-convex on a real interval ,ab if for all ,, yab and 0,1 , we have 1 1 xyfxf y. (1.3) If the above inequality reversed, then f is termed log-concave. We define for ,0xy , ln ln , , xy y xy Lxy xy In [2] the following result is achieved: Theorem 1.1. Let f be a positive log-convex function on [a, b]. The n 1d, b a atLfafa ba . (1.4) For f a positive log-concave function, the inequa lity is reversed. 2. Lemmas The following lemmas are needed for our aim. Lemma 2.1. Let 0t1, then the following ine- quality holds 1 11 t t tt 2. (2.1) Proof. Set 1 ln1ln1 2. t t ftt t We have lnln10,ft tt for 12.t 11 0. 1 ft tt Therefore f attains its minimum at 12t which is 12 Hence 0ft which implies , and (2.1) follows. 0 ft e Lemma 2.2. For 0,0ab t1, the following inequality holds 1 1 ,12 ,12 tt tt abt ab abt , (2.2) and for , the following inequality holds 0, 0, 01ab t 11 2. tt tt aba babab (2.3)
 W. T. SULAIMAN 900 Proof. For 12,t we have 1/2 t ba ba, which implies 1tt aba b , and for 12,t 1/2 , t ab ba which implies 1. tt abab We also have 2 11 2222 0 tt tt abab , which implies 11 2tt tt aba bab . Set Then, on keeping b fixed, we have 11 . tt tt fta babab 11 21 1 110, 11. tt tt tt tt fatabtabforab fattabt ttab . , As , f attains its mini- mum at which is 0, therefore and (2.3) is satisfied. 1 21 0 ab fat ta ab 0,fa Although some of the coming results (Lemma 2.3 and theorem 3.1) are known, but we prove them by new sim- ple method. Lemma 2.3. Let then the following inequal- ity holds ,0ab , ln ln2 ab ab ab ab (2.4) Proof. Left inequality. Let us assume that Set .ba 1212 ln ,1.fxxxxx (2.5) 1232 1 11 0 22 fx xx x as 2 14 3412 321 02 xxx x . Therefore f is non-decreasing, and that implies x . The result follows by putting 10f ba in (2.5). Right inequality. Let and let ,ab. ab Set 1 ln 2,1. 1 fx xx (2.6) We have 2 140 1 fx xx as 2 2 14 10 1 xxx . Then f is non-decreasing, and hence 10.fx f The result follows by putting ab in (2.6). Lemma 2.4. The function 1, ln 1 xx (2.7) is non-decreasing. Proof. 1 22 2 ln 1. ln ln 110, x xx fx x gx xx therefore g is non-decreasing. Since 10g, then 0,gx and hence 0,fx that is f is non-de- creasing. 3. Theorems Theorem 3.1. Let f be a positive log-convex function on [a,b], then f satisfies (1.1). Proof. This can be achieved immediately as the log-convex function is convex which follows from the fact that “Every increasing convex function of a convex function is convex” which implies that ln x fx e is convex. Or the proof can be achieved by following the definition: Making use of lemma 2.2, we have 12 12 12 12 11 00 11 1 1 0 0 11 dd 22 1dd 11 1 d1d1d 22 dd 22 bb aa bb aa b a tt tt ab abxx ffxfabxfxx ba ba fabxxfx x ba f xxftatbtftatb t ba . 2 afbfaf bfafbfafb tt The following giving a refinement to theorem 3.1. Theorem 3.2. Let f be a log convex function. Then the following inequality holds 2 11 dd 2lnln bb aa 2 bfafafb ab ffxxfxx babafb fa (3.1) Proof. Copyright © 2011 SciRes. AM
 W. T. SULAIMAN901 1/ 21/ 2 11 dd 22 2 11 ()dd dd, bb aa bbb aaa ab ababxx ffxfx ba ba 1 b a abx fxxfabxxfxxfxx ba baba which implies 2 2 11 dd 2ln bb aa ln2 bfafafb ab ffxxfxx bafb fa ba in view of [2] and Lemma 2.3. The following presents an extension to Fejer’s gener- alization (1.2) for log-convex functions Theorem 3.3. Let f be log convex, g is positive, inte- grable and symmetric to 2xab Then the fol- lowing inequality holds 22 ddd 2 dd ln ln2 bb b aa a bb aa ab fgxxfxgxxfxgxx fb fafafb bagxx bagxx fb fa . (3.2) Proof. 1/2 1/2 12 12 1/21/21/2 1/2 12 12 dd()d 22 () ddd ddd d bb b aa a bbb aaa bbbb aaaa ab abxx fgxxfgxxfabxfx gxx fabxgabx fxgxxfabxgabxxfxgxx f xgxxfxgxxxf xgxx 12 12 ba which implies 22 dd 2 bb b aa a ab d gx xfxgxxbafxgxx Now, for 012t , we have 11 1 00 1 0 d11d1d ()()1dd, ln ln btt a b a fxgxxbafta tbgtatbtbafafbgtatbt fb fa bafafbgtatbtgxx fb fa in view of Lemmas 2.2 and 2.3. Also, we have for 12,t 11 11 00 1 0 d11d 1 1dd, ln ln btt a b a fxgxxbaftatbgtatbtbafafbgtatbt fb fa bafafb gtatbtgxx fb fa d Copyright © 2011 SciRes. AM
 W. T. SULAIMAN 902 in view of Lemmas 2.2 and 2.3. Consequently, we obtain, by Lemma 2.2, dd ln ln2 bb aa fb fafafbd. b a xgxxgxx gxx fb fa This completes the proof of the theorem. The following is another refinement of theorem 3.1. Theorem 3.4. Assume that be an increas- ing log-convex function. Then for all :fI 0,t1, we have ,d 2lnln b a , 2 afb fafb ab fwabfxxWt fa fb (3.3) where 3 , 44 aba b wab ff 3 , (3.4) 11 1. ln 1 lnlnln 1 fta tbfafbftatb Wt tt tatbfafb ftatb (3.5) Proof. We have via Lemmas 2.3 and 2.4 2 2 (1 ) 2 (1 ) 2 1313332 ()d d 224244 4 211 1 ddd()dd 2 1 (1)( ) 1 ab b ab a ab tat b bb b ab aaatatb ababa baba b fffwtfx ba fx xfxxfx xfxxfx x baba ba tfx tba xfxx (1 ) (1 ) 1 dd () 1() 1 (1)( ) ln1ln()lnln1 () (1 ). ln ln()ln lnln ln2 tat bb atatb xt fxx tb a ftatbfafbfta tb ttWt ftatbfafbftatb fb fafb fafb fafa fb tt fb fafb fafb fa Theorem 3.5. Let f is log-convex and g is non-negative, integrable, 1111,ppq, then the following inequality holds 11 dd ln ln pq pp bb q aa fa fb ba xgx xgx x pfa fb . (3.6) Proof. We have, via Holder’s inequality 1 111 1 0 11 1 1 (1 ) 0 0 ddd1d d ddd()d q ppq bbb b pq pq aaa a pt pq bb ptptqpq aa fxgxxfxxgxxba f tatbtgxx fa bafafbtg xxbafbtgxx fb 1 11 ()d ln ln q pq pp bq a fa fb ba gxx pfa fb Copyright © 2011 SciRes. AM
 W. T. SULAIMAN903 4. References r die Fourierreihen, II,” Math. Naturwiss Anz. Ungar. Akad. Wiss, Hungarian, Vol. 24, 1906, pp. 369-390. [2] P. M. Gill, C. E. M. Pearse and J. Pečarić, “Hadamard’s for R-Convex Functions,” Mathematical Ine- [1] L. Fejér, “UbeInequality qualities & Applications, Vol. 215, No. 2, 1997, pp. 461-470. Copyright © 2011 SciRes. AM
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