Florida International University FIU Digital Commons Department of Mathematics and Statistics College of Arts, Sciences & Education 200 A Generalization of Bernoulli's Inequality Laura De Carli Department of Mathematics and Statistics, Florida International University, decarlil@fiu.edu Steve M. Hudson Department of Mathematics and Statistics, Florida International University, hudsons@fiu.edu Follow this and additional works at: http://digitalcommons.fiu.edu/math_fac Part of the Mathematics Commons Recommended Citation DE CARLI, Laura; HUDSON, Steve M.. A generalization of Bernoulli's inequality. Le Matematiche, [S.l.], v. 65, n., p. 09-7, dec. 200. ISSN 2037-5298. Available at: This work is brought to you for free and open access by the College of Arts, Sciences & Education at FIU Digital Commons. It has been accepted for inclusion in Department of Mathematics and Statistics by an authorized administrator of FIU Digital Commons. For more information, please contact dcc@fiu.edu.
LE MATEMATICHE Vol. LXV 200) Fasc. I, pp. 09 7 doi: 0.448/200.65..9 A GENERALIZATION OF BERNOULLI S INEQUALITY LAURA DE CARLI - STEVE M. HUDSON We prove the following generalization of Bernoulli s inequality c k + a jk )) s c k + sa jk ) where 0 s, under suitable conditions on the a jk and the c k. We also prove the opposite inequality when s. These inequalities can be applied to Weierstrass product inequalities.. Introduction The classical Bernoulli inequality is + x) s + sx ) for x > and 0 s. For s >, the inequality reverses. This has been generalized in a number of ways. See Mitrinović and Pećarić [5] for a survey. The version in this paper, Theorem. below, can be expressed in terms of matrices. It is more general than the versions in [7] and [8]. See also [3]. Entrato in redazione: 4 maggio 200 AMS 2000 Subject Classification: 26D5. eywords: Bernoulli s inequality, Weierstrass product inequalities.
0 LAURA DE CARLI - STEVE M. HUDSON Theorem.. Let A = a j k ) be a real matrix, with a j k > for all j and k. Let c k > 0 for all k, and let S = k= c k. Assume m + a jk) M. Consider the following conditions on s and A: C) 0 s, and S s + M) s + sm, 2) C2) 0 s, c, and for all j, k, a j k a j. C3) s > and a j k > s for all j, k, and S s + m) s + sm, 3) C4) s >, c, and for all j, k, s a j k a j. Define LA) = c k + a jk )) s and RA) = c k + sa jk ). Then if C) or C2) hold, LA) RA) while if C3) or C4) hold, LA) RA). We require and to be finite for simplicity; the theorem can easily be extended to the infinite case by taking limits. A special case worth mentioning is when =, c = and 0 s. The inequality LA) RA) reduces to s + a j )) + sa j ) 4) which is a straightforward consequence of ). Observe that when s >, the condition a j > s is necessary for LA) RA) or even the reverse of 4)). Without this condition, the product on the right hand side might include an even number of large negative factors. Here is an example of how our results can be useful in a perhaps surprising manner. Example. If 0 < s <, the following inequality holds + 0 ) sinx s x dx sin sx) + dx. 5) 0 sx
A GENERALIZATION OF BERNOULLI S INEQUALITY ) Recalling that, for x [ π,π], sinx x = x2 see e.g. []) we can π j) 2 prove 5) by applying the theorem to the Riemann sums of these integrals. That is, given 0 s and a regular partition {x k } k=,..., of [0,], let a jk = x2 k, π j) 2 for j,k and let a j0 = 0 for simplicity, we may start with k = 0 instead of k = ). Let c k = for k and let c 0 =, so C2) holds. Then + ) ) s x2 k π j) 2 = LA) RA) = + ) sx2 k π j) 2, and the inequality 5) follows by letting and go to infinity. We could prove a slightly more general version of 5) by replacing [0,] with [a,b], where π < a < b < π. In section 2, we show that conditions like the Ci) above are necessary for universal comparability of LA) and RA), and then we prove our main results. In section 3, we apply our generalized Bernoulli inequalities to prove new Weierstrass product inequalities. Acknowledgements. We wish to thank the referee for his/her many suggestions and comments which helped us to improve our paper. 2. Remarks on the Ci) and the proof of Theorem. We cannot suggest any simple necessary and sufficient conditions for the inequality LA) RA) in Theorem., but will show that with weaker conditions than the given Ci) it can fail. For example, consider C5) 0 s, and < a jk for all j, k. C6) s, and s < a jk for all j, k. The following example shows C5) cannot replace C) or C2) in the theorem. Let s = /2 and let c k = / for all k. Let a j = for all j and let a jk = 0 whenever k 2. So, C5) holds, but neither C) nor C2) do. The product in LA) is 2 when k =, and otherwise is. The product in RA) is 3/2) when k =, and otherwise is. So, LA) = /2 [2 + )] /2, RA) = [3/2) + )]. After multiplying both sides by = 4/3) or the nearest integer) LA) 2 2 /2 = 8/3) /2,
2 LAURA DE CARLI - STEVE M. HUDSON while RA) 3/2) + 4/3) 23/2). Since 8/3) /2 > 3/2, LA) > RA) for large enough, so we have an example that shows that C5) is insufficient for LA) RA). We now show that if C3) and C4) are replaced by the weaker C6), then LA) RA) is not necessarily true. Set s = 2, = 2, and a j = 0, and a j2 = for all j s. We also let c 2 = 2 and c = 2 with 4. Then, LA) = 2 2 ) 2 < 4, and RA) = 2 + 2) + 2 ) > 3/2) > 4 > LA). The a jk in the examples above spike as a function of k; that is, there is a value of k for which a jk is much larger than the average. M is much bigger than m. The conditions C)-C4) can be viewed as anti-spiking conditions. For example, with spiking 2) in C) is a fairly strong condition on s and S. But when S = and m = M, 2 is just the standard Bernoulli inequality; and 3) is similar. So, C)-C4) cannot be replaced by the simpler C5) or C6), but it is possible that they can be weakened in other ways. For example, in the special case, =, 0 s, and S = c k =, the inequality LA) RA) holds without any anti-spiking assumptions. Indeed, s ) s c k + a k )) = + c k a k + s c k a k = c k + sa k ). 6) Lemma 2. below reduces the proof of Theorem. to the special case in which the a jk do not depend on j. Assume the s and A = a jk ) satisfy C5) or C6). Define p k = + a jk) /. Recall that in Theorem. we have assumed m p k M. Without loss of generality we can assume M = max k p k and m = min k p k. Replacing each a jk by p k defines a new matrix P with LP) = LA). Also, if A satisfies any of the conditions Ci i 4) in Theorem., then P does too. Lemma 2.. With P as above: a) If A satisfies C5), then RP) RA). b) If A satisfies C6), then RP) RA). Proof. We consider only case a), since the proof in case b) is similar. So, 0 < s and < a jk. Consider the problem of finding a matrix B = b jk ) such that + b jk ) = + a jk )
A GENERALIZATION OF BERNOULLI S INEQUALITY 3 for all k, which minimizes RB). For compactness, we also require that min a jk b jk max a jk for all j,k,which insures that a solution B exists, with RB) RA). We will show that b ik = b jk for all i, j,k, which implies B = P and proves the Lemma. If not, then without loss of generality, i and j are and 2, k = and b < b 2. Define β by + β) 2 = + b ) + b 2 ). Define a new matrix B from B by replacing both b and b 2 with β. This substitution does not change + b j), so B is admissible for the minimization problem above. Also, the variables b and b 2 only occur in one term of RB): + sb ) + sb 2 ) = [ s] + s[ + b ])[ s] + s[ + b 2 ]). So, for some positive constant c, RB) RB) = c[ + b )] + [ + b 2 ] 2[ + β]) = c[ + b ] /2 [ + b 2 ] /2 ) 2 > 0. This contradicts our assumption that B minimizes R and completes the proof of the Lemma. We now proceed with the proof of Theorem.. Proof that C) implies LA) RA). With P defined as above, LA) = LP), and by Lemma 2., RP) RA). Then, LP) = ) s c k + p k ) ) s c k + M) = S s + M) s S + ms) c k + sp k ) = RP), proving LA) RA). The proof that C3) implies LA) RA) is similar. Proof that C2) implies LA) RA). By Lemma 2., it suffices to prove that LP) RP). By C) and the definition of p k, < p k p, for all k; also, c. The calculation below uses the facts that s a s is an increasing function of s when a > and is decreasing when a < in the first two inequalities, and Bernoulli in the last one: LP) = ) s c k + p k )
4 LAURA DE CARLI - STEVE M. HUDSON = c s + p ) s + c s + p ) s + c s + p ) s + = c s + p ) s + c + sp ) + k=2 k=2 k=2 k=2 k=2 c c k c c k c c k ) ) + s pk + p ) ) + pk + p ) ) + s pk + p c s c k + p k ) s c k + sp k ) = RP). and the proof is complete. The proof that C4) implies LA) RA) is similar. We conclude this section with a proposition which is very similar in spirit to Lemma 2.. That Lemma showed that RA) is minimal subject to certain constraints) when A = P; that is, when the a jk depend only on k. Our next proposition goes further; if L = LB) is fixed, then the minimal RB) occurs when b jk = b k has only 3 distinct values. In effect, it shows we may assume 3. Though not used in this paper, we believe that this second reduction may have independent interest, and might be applied to other generalized Bernoulli inequalities. Proposition 2.2. Fix 0 s, {c k },,, L and < m < M. Consider the family of all matrices B with constant columns determined by the b k, and so that m b k M for k =,..., and LB) = L. If such B minimizes RB), then all b k, with the possible exception of one, are equal to m or M. This conclusion is also true in the opposite case; when s >, m > s and RB) is maximal. Proof. Suppose first that 0 < s <. We will show that one element in every pair of b k s is either m or M. We may take the pair to be {b,b 2 } and can assume b b 2. We will treat these elements as variables in the interval [m,m] and consider minimizing RB). Let t = + b ) and t 2 = + b 2 ), with + m) t t 2 + M). Fix c t + c 2 t 2 = T, so that changing the t i will not affect LB). We will show that RB) is minimal when at least one of the t i is an endpoint, meaning that one of the b i is an endpoint. The summand in RB) affected by t i is equal to st i + s). Thus, we minimize f t ) = c st + s) + c 2 st 2 + s). where t 2 is a function of t. Note that dt 2 dt = c c 2. So,
A GENERALIZATION OF BERNOULLI S INEQUALITY 5 d f t ) = sc st dt + s) t sc st 2 + s) t 2 = sc [ + sb ) + b ) + sb 2 ) + b 2 ) ] It is easy to check that, for s <, +sx +x decreases with x. So, f > 0, which implies the minimum of f occurs when t is minimal or when t 2 is maximal, and we are done. When s >, +sx +x increases with x, so f < 0, and the maximum of f occurs when t is minimal or t 2 is maximal. 3. Weierstrass Inequalities We can use Theorem. to prove some Weierstrass product inequalities. In order to study the convergence of infinite products, it is useful to find lower bounds for products of the form n i= +x i) and n i= x i) which are named after. Weierstrass, in terms of linear functions.. Weierstrass was probably the first to prove the following inequalities: and if 0 x + n i= x i n i= n i= + x i ), x i >, 7) x i n i= x i ). 8) These inequalities and their generalizations have attracted a lot of interest. See for example [2], [4], [9] just to cite a few. The following is a corollary of Theorem.: Theorem 3.. Let c k 0 and 0 a jk < and 0 < s <. Then, s + c k a jk )) + c k sa jk ). 9) When s > and 0 a jk s the inequality reverses. Proof. In Theorem., set c = and a j = 0 for all j. For k 2, replace a jk by a j,k, with a similar re-indexing of the c k. Then C2) is satisfied and Theorem. gives immediately Theorem 3.. For the reverse, apply C4). Theorem 3.2. Let a jk [ s,] and c k 0. Assume c k = and s and m + a jk) / M and + sm + m) s.
6 LAURA DE CARLI - STEVE M. HUDSON Then + s c k a jk j c k Proof. By C3) of Theorem., and 7), s c k + a jk )) c k + a jk )) s. + sa jk ) as required. c k + s j ) a jk = + s c k a jk j Theorem 3.3. Let s and let a jk [0,/s]; Let c k 0. Then s 2 s c k a jk + c k a jk )). j Proof. Follows from Theorem 3. and 8). Indeed, s + c k a jk )) + + c k s j c k ) a jk = 2 s c k a jk. j sa jk ) REFERENCES [] L. V. Ahlfords, Complex analysis. McGraw-Hill, 979. [2] H. Alzer, Inequalities for Weierstrass products. Portugal. Math. 47 ) 990), 47 59. [3] L. Gerber, An extension of Bernoulli s inequality. Amer. Math. Monthly 75 968), 875 876. [4] M. S. lamkin, Extensions of the Weierstrass product inequalities. II, Amer. Math. Monthly 82 7) 975), 74 742.
A GENERALIZATION OF BERNOULLI S INEQUALITY 7 [5] D. S. Mitrinović -. E. Pećarić, Bernoulli s inequality. Rend. Circ. Mat. Palermo II Ser. 42 3) 993), 37-337. [6] D. S. Mitrinović - P. M. Vaic, Analytic Inequalities. Berlin-Heidelberg-New York, 970. [7] Y. Ma - L. Zhang, Some research for Bernoulli s inequality. Int.. Appl. Math. Stat. 7) 2007), 203 204. [8] Huan-Nan Shi, Generalizations of Bernoulli s inequality with applications.. Math. Inequal. 2 ) 2008), 0 07. [9] S. Wu, Some results on extending and sharpening the Weierstrass product inequalities.. Math. Anal. Appl. 308 2) 2005), 689 702. LAURA DE CARLI Dept. of Mathematics Florida International Univ. Miami, FL 3399 U.S.A. e-mail: decarlil@fiu.edu STEVE M. HUDSON Dept. of Mathematics Florida International Univ. Miami, FL 3399 U.S.A. e-mail: hudsons@fiu.edu