aa r X i v : . [ m a t h . C A ] J un FINITE SECTIONS OF WEIGHTED CARLEMAN’S INEQUALITY
PENG GAO
Abstract.
We study finite sections of weighted Carleman’s inequality following the approach ofDe Bruijn. Similar to the unweighted case, we obtain an asymptotic expression for the optimalconstant. Introduction
The well-known Carleman’s inequality asserts that for convergent infinite series P a n with non-negative terms, one has ∞ X n =1 ( n Y k =1 a k ) n ≤ e ∞ X n =1 a n , with the constant e best possible.There is a rich literature on many different proofs of Carleman’s inequality as well as its gener-alizations and extensions. We shall refer the readers to the survey articles [7] and [5] as well as thereferences therein for an account of Carleman’s inequality.From now on we will assume a n ≥ n ≥ ∞ X n =1 G n ≤ U ∞ X n =1 a n , where G n = n Y k =1 a λ k / Λ n k , Λ n = n X k =1 λ k , λ k ≥ , λ > . Using Carleman’s original approach in [2], the author [4] proved the following:
Theorem 1.1.
Suppose that (1.2) M = sup n Λ n λ n log (cid:16) Λ n +1 /λ n +1 Λ n /λ n (cid:17) < + ∞ , then inequality (1.1) holds with U = e M . In this paper, we consider finite sections of weighted Carleman’s inequality (1.1):(1.3) N X n =1 G n ≤ µ N N X n =1 a n . where N ≥ λ k = 1 (the unweighted case), De Bruijn [3] had shownthat the best constant satisfies µ N = e − π e (log N ) + O (cid:16) N ) (cid:17) . Date : June 30, 2007.2000
Mathematics Subject Classification.
Primary 26D15.
Key words and phrases.
Carleman’s inequality.
It is our goal in this paper to obtain similar asymptotic expressions for µ N for the weightedCarleman’s inequality following De Bruijn’s approach in [3]. We shall prove the following Theorem 1.2.
Assume (1.2) holds with { λ k } ∞ k =1 a non-decreasing sequence satisfying sup k λ k +1 λ k < + ∞ , (1.4) M + log( λ k /λ k +1 ) ≤ Λ k +1 λ k log (cid:16) Λ k +1 /λ k +1 Λ k /λ k (cid:17) , (1.5) Λ k λ k log (cid:16) Λ k +1 /λ k +1 Λ k /λ k (cid:17) = M + O (cid:16) λ k Λ k (cid:17) , (1.6) λ k Λ k = Ck + O ( 1 k ) , C > , (1.7) inf k (cid:16) Λ k +1 λ k +1 − Λ k λ k (cid:17) > . (1.8) Then for any integer N ≥ , inequality (1.3) holds with the best constant satisfying: µ N = e M − π e M C (log N ) + O (cid:16) N ) (cid:17) . We note here that (1.8) implies
M >
0, which we shall use without further mentioning throughoutthe paper. We may also assume N ≥ Preliminary Treatment
It is our goal in this section to give an upper bound for the number U N appearing in (1.3).We first recall the author’s approach in [4] (following that of Carleman in [2]) for determining themaximum value µ N of P Nn =1 G n in (1.3) subject to the constraint P Nn =1 a n = 1 using Lagrangemultipliers. It is easy to see that we may assume a n > ≤ n ≤ N when the maximum isreached. We now define F ( a ; µ ) = N X n =1 G n − µ ( N X n =1 a n − , where a = ( a n ) ≤ n ≤ N . By the Lagrange method, we have to solve ∇ F = 0, or the following systemof equations:(2.1) µa k = N X n = k λ k G n Λ n , ≤ k ≤ N ; N X n =1 a n = 1 . We note that on summing over 1 ≤ k ≤ N of the first N equations above, we get N X n =1 G n = µ. Hence we have µ = µ N in this case which allows us to recast the equations (2.1) as: µ N a k λ k = N X n = k G n Λ n , ≤ k ≤ N ; N X n =1 a n = 1 . On subtracting consecutive equations, we can rewrite the above system of equations as: µ N ( a k λ k − a k +1 λ k +1 ) = G k Λ k , ≤ k ≤ N − µ N a N λ N = G N Λ N ; N X n =1 a n = 1 . INITE SECTIONS OF WEIGHTED CARLEMAN’S INEQUALITY 3
Now following the notations in [3], we define for 1 ≤ k ≤ N − h k = log G k a k , so that we can obtain a recursion expressing h k +1 in terms of h k as follows: h k +1 = Λ k Λ k +1 h k − Λ k Λ k +1 log (cid:16) λ k +1 λ k − λ k +1 Λ k µ N e h k (cid:17) , ≤ k ≤ N − . We now define a sequence of real functions h k ( µ ) inductively by setting h ( µ ) = 0 and(2.2) h k +1 ( µ ) = Λ k Λ k +1 h k ( µ ) − Λ k Λ k +1 log (cid:16) λ k +1 λ k − λ k +1 Λ k µ e h k ( µ ) (cid:17) , ≤ k ≤ N − . We note that h k ( µ N ) = h k for 1 ≤ k ≤ N − h N ( µ N ) = Λ N − Λ N h N − ( µ N ) − Λ N − Λ N log (cid:16) λ N λ N − − λ N Λ N − µ N e h N − ( µ N ) (cid:17) = Λ N − Λ N log (cid:16) G N − a N − (cid:17) − Λ N − Λ N log (cid:16) λ N λ N − − λ N Λ N − µ N (cid:16) µ N (cid:16) Λ N − λ N − − Λ N − λ N a N a N − (cid:17)(cid:17)(cid:17) = Λ N − Λ N log (cid:16) G N − a N (cid:17) = log (cid:16) G N a N (cid:17) = log( µ N Λ N λ N ) . We now show by induction that if µ ≥ e M , then for any 2 ≤ k ≤ N ,(2.3) h k ( µ ) ≤ M Λ k − Λ k . As we have seen above that h N ( µ N ) = log( µ N Λ N /λ N ) ≥ log µ N ≥ M when µ n ≥ e M , this forces µ N < e M .Now, to establish (2.3), we first consider the case k = 2. As h = 0, We have by (2.2),(2.4) h ( µ ) = − Λ Λ log (cid:16) λ λ − λ Λ µ (cid:17) . It is easy to see that h ( µ ) ≤ M Λ / Λ is equivalent to λ λ e M ≥ λ Λ e M µ + 1 . As e M /µ ≤
1, the above inequality follows easily from the assumption (1.2). Now assume inequality(2.3) holds for k ≥
2, then by (2.2) again, it is easy to see that for (2.3) to hold for k + 1, it sufficesto show that λ k +1 λ k e Mλ k / Λ k ≥ λ k +1 Λ k e M µ + 1 , and this again follows easily from the assumption (1.2).3. The Breakdown Index
As in [3], we now try to evaluate h k ( µ ) consecutively from (2.2) for any µ >
0, starting with h = 0. Certainly we are only interested in the real values of h k and hence we say that the procedurebreaks down at the first k where λ k +1 /λ k − λ k +1 / (Λ k µ ) e h k ( µ ) ≤
0, or equivalently,(3.1) h k ( µ ) ≥ log( µ Λ k /λ k ) . PENG GAO
We define the breakdown index N µ as the smallest k for which inequality (3.1) holds if there issuch a k and we put N µ = + ∞ otherwise. Thus for all µ > h k ( µ ) is defined forall k ≤ N µ .Note that (2.3) implies N µ = + ∞ when µ ≥ e M . So from now on we may assume 0 < µ < e M and it is convenient to have some monotonicity properties available in this case. We have h ( µ ) = 0for 0 < µ < e M and we let µ be the largest µ for which inequality (3.1) holds for k = 1, thisimplies µ = 1. Now h ( µ ) is defined for µ > µ , and h ( µ ) is given by (2.4), which is a decreasingfunction of µ for µ > µ . Note also that the right-hand side expression of inequality (3.1) is anincreasing function of µ for any fixed k . It follows thatlim µ → µ +1 h ( µ ) = + ∞ ; h ( e M ) ≤ M (1 − λ / Λ ) < log( e M Λ /λ ) ≤ log( µ Λ /λ ) . Thus there is exactly one value of µ < e M for which inequality (3.1) holds with equality for k = 2and we define this value of µ to be µ . This procedure can be continued. At each step we arguethat h k ( µ ) is defined and decreasing for µ > µ k − , thatlim µ → µ + k − h k ( µ ) = + ∞ ; h k ( e M ) ≤ M (1 − λ k / Λ k ) < log( e M Λ k /λ k ) ≤ log( µ Λ k /λ k ) . We then infer that µ k is uniquely determined by h k ( µ ) = log( µ Λ k /λ k ). Moreover, h k +1 ( µ ) is againdefined and decreasing for µ > µ k as both terms on the right of (2.2) are decreasing functions of µ .Thus by induction we obtain that(3.2) 1 = µ < µ < µ < . . . < e M , and that h k +1 ( µ ) is defined and decreasing for µ > µ k . Moreover, h k ( µ ) > log( µ Λ k /λ k ) if µ k − <µ < µ k , h k ( µ k ) = log( µ k Λ k /λ k ), h k ( µ ) < log( µ k Λ k /λ k ) if µ > µ k .It follows that the breakdown index N µ equals 1 if µ ≤ µ , 2 if µ < µ ≤ µ , etc. We remarkhere that for fixed µ ≤ e M , the h k ( µ )’s are non-negative and increase as k increases from 1 to N k .This follows from (2.2) by noting that(3.3) h k +1 ( µ ) − h k ( µ ) = − λ k Λ k +1 h k ( µ ) − Λ k Λ k +1 log (cid:16) λ k +1 λ k − λ k +1 Λ k µ e h k ( µ ) (cid:17) . It thus suffices to show the right-hand side expression above is non-negative. Equivalently, this is λ k +1 /λ k ≤ f ( h k ( µ )), where f ( x ) = λ k +1 Λ k µ e x + e − λk Λ k x . It is easy to see that f ( x ) is minimized at x = Λ k / Λ k +1 log( λ k µ/λ k +1 ). Note also that λ k +1 Λ k µ e x = λ k Λ k e − λk Λ k x . It follows that f ( x ) ≥ f ( x ) = Λ k +1 Λ k e − λk Λ k x . It follows from (1.5) thatlog( λ k e M /λ k +1 ) = M + log( λ k /λ k +1 ) ≤ Λ k +1 λ k log (cid:16) Λ k +1 /λ k +1 Λ k /λ k (cid:17) . It is easy to see that the above inequality implies that f ( x ) ≥ λ k +1 /λ k so that the h k ( µ )’s increaseas k increases from 1 to N k .The breakdown condition (3.1) is slightly awkward. We now replace it by a simpler one, forexample, h k > max(2 , M ), by virtue of the following argument. Let 0 < µ < e M and assume that N is such that h N > max(2 , M ). Note that (1.7) implies that lim k → + ∞ Λ k /λ k = + ∞ so that INITE SECTIONS OF WEIGHTED CARLEMAN’S INEQUALITY 5 the right-hand side expression of (3.1) approaches + ∞ as k tends to + ∞ . Hence we may assume N µ ≥ N without loss of generality. Then we havelog N µ − log N = O (1) . For, if N ≤ k ≤ N µ , the right-hand side of (3.3) equals − λ k Λ k +1 h k ( µ ) − Λ k Λ k +1 log (cid:16) λ k +1 λ k (cid:17) − Λ k Λ k +1 log (cid:16) − λ k Λ k µ e h k ( µ ) (cid:17) = − Λ k Λ k +1 log (cid:16) λ k +1 λ k (cid:17) + λ k Λ k +1 (cid:16) e h k ( µ ) µ − h k ( µ ) (cid:17) + Λ k Λ k +1 + ∞ X i =2 i (cid:16) λ k Λ k e h k ( µ ) µ (cid:17) i (3.4) ≥ λ k Λ k +1 (cid:16) e h k ( µ ) µ − h k ( µ ) − Λ k λ k log (cid:16) λ k +1 λ k (cid:17)(cid:17) . Note that, in view of (1.2) and (1.6),(3.5) Λ k λ k log (cid:16) λ k +1 λ k (cid:17) = − Λ k λ k log (cid:16) Λ k +1 /λ k +1 Λ k /λ k (cid:17) + Λ k λ k log (cid:16) Λ k +1 Λ k (cid:17) = 1 − M + O (cid:16) λ k Λ k (cid:17) . As ( e h − M − h + M − h − increases for h ≥ max(2 , M ), we conclude that there exists a constant C > N independent of µ such that for k ≥ N , h k +1 ( µ ) − h k ( µ ) ≥ λ k Λ k +1 (cid:16) e h k ( µ ) − M − h k ( µ ) + M − (cid:17) + O (cid:16) λ k Λ k Λ k +1 (cid:17) > C λ k Λ k +1 h k ( µ ) . We may assume N ≥ N from now on without loss of generality and we now simply the aboverelations by defining d N , d N +1 , . . . , starting with d N = h N , and(3.6) d k +1 − d k = C λ k Λ k +1 d k . Obviously we have d k ≤ h k ≤ log( µ Λ k /λ k ) for N ≤ k ≤ N µ . We use the boundlog( µ Λ k /λ k ) ≤ M + log(Λ k /λ k ) ≤ M − k /λ k ≤ M Λ k /λ k . to get that d k ≤ M Λ k /λ k for N ≤ k ≤ N µ . It follows from (3.6) that d k +1 − d k ≤ C M d k . The above implies that we have d k +1 ≤ ( C M + 1) d k for N ≤ k ≤ N µ and (3.6) further impliesthat(3.7) d k +1 − d k ≥ C λ k ( C M + 1)Λ k +1 d k d k +1 . We now apply (1.7) to obtain via (3.7) that there exists a constant C > N independent of µ such that for k ≥ N , d − k − d − k +1 ≥ C k + 1 . Certainly we may assume N ≥ N as well. Summing the above for N ≤ k ≤ N µ − , M ) ≥ d − N ≥ X N ≤ k ≤ N µ − C k + 1 . It follows from this that(3.8) log N µ − log N = − X N ≤ k ≤ N µ − log( kk + 1 ) ≤ X N ≤ k ≤ N µ − k + 1 + O (1) = O (1) . PENG GAO
We shall see in what follows that the relation (3.8) implies that there is no harm studyinglog N in stead of log N µ . So from now on we shall concentrate on finding the smallest k such that h k ( µ ) > max(2 , M ). 4. Heuristic Treatment
Our problem is, roughly, to determine how many steps we have to take in our recurrence (3.3)in order to push h k beyond the value of max(2 , M ), assuming that µ is fixed, µ < e M and µ closeto e M . Now assume we are able to neglect all the other terms of the right-hand side expression in(3.4) other than the first two terms, then we have a recurrence which can be written as∆ h = λ k Λ k +1 (cid:16) e h k ( µ ) µ − h k ( µ ) − Λ k λ k log (cid:16) λ k +1 λ k (cid:17)(cid:17) . In view of (3.5), we may replace the last term above by 1 − M and we may further consider thefollowing recurrence using (1.7):∆ h = Ck + 1 (cid:16) e h k ( µ ) µ − h k ( µ ) + M − (cid:17) . Next we consider k as a continuous variable, and we replace the above by the correspondingdifferential equation, that is, we replace ∆ h by dh/dk . Then we get d log( k + 1) dh = C − (cid:16) µ − e h − h + M − (cid:17) − . This suggests that if N is the number of steps necessary to increase h from 0 to about max(2 , M ),then log N is roughly equal to(4.1) 1 C Z max(2 , M )0 dhµ − e h − h + M − . The integrand has its maximum at h = log µ , and this is close to M . In the neighborhood of thatmaximum it can be approximated by12 ( h − log µ ) + M − log µ. Therefore the value of (4.1) can be compared with1 C Z + ∞−∞ dh ( h − log µ ) + M − log µ = √ πC (cid:16) log( e M /µ ) (cid:17) − / . From this we see that for µ < e M , µ → e M , we expect to have(4.2) log N µ = √ πC (cid:16) log( e M /µ ) (cid:17) − / + O (1) . From this we see that if µ → e M , then log N µ tends to infinity. This also implies that for thesequence { µ k } defined as in (3.2), one must have lim k → + ∞ µ k = e M . For otherwise, the sequence { µ k } is bounded above by a constant < e M and on taking any µ greater than this constant (andless than e M ), then the left-hand side of (4.2) becomes infinity (by our definition of N µ ) but theright-hand side of (4.2) stays bounded, a contradiction.Note that if µ = µ N , then N µ = N , it follows from (4.2) thatlog( e M /µ N ) = 2 π C (cid:16) log N + O (1) (cid:17) − . It is easy to see that the above leads to the following asymptotic expression for µ N : µ N = e M − π e M C (log N ) + O (cid:16) N ) (cid:17) . INITE SECTIONS OF WEIGHTED CARLEMAN’S INEQUALITY 7
There are various doubtful steps in our argument above, but the only one that presents a seriousdifficulty is the omitting of all the other terms of the right-hand side expression of (3.4). Certainlythose terms can be expected to give only a small contribution if k is large but the question iswhether this contribution is small compared to µ − e h − h + M −
1. The latter expression can besmall if both h k − M and µ − e M are small, and it is especially in that region that the integrandof (4.1) produces its maximal effect. 5. Lemmas
Lemma 5.1.
For any given number η > , < ǫ < M , one can find an integer k > η and anumber β , e M − < β < e M such that for β < µ ≤ e M , (5.1) M − ǫ < h k ( µ ) < log µ − M λ k Λ k . Proof.
Note first that by (2.3) and our discussions in Section 3 that the h k ( e M )’s are non-negative,we have 0 ≤ h k ( e M ) ≤ M Λ k − Λ k . Let k be an integer so that for all k ≥ k , M Λ k − Λ k > M − ǫ. We may assume that k ≥ k from now on and note that not all h k ( e M ) are ≤ M − ǫ . Otherwise,it follows from (3.3), (1.7), (3.4) and (3.5) that h k +1 ( e M ) − h k ( e M ) ≥ λ k Λ k +1 (cid:16) e h k ( e M ) e M − h k ( e M ) + M − (cid:17) + O ( 1 k ) . Note that if h k ( e M ) ≤ M − ǫ then e h k ( e M ) e M − h k ( e M ) + M − ≥ e M − ǫ − M − M + ǫ + M − > . It follows from (1.7) and the fact that P ∞ k = k ( k + 1) − = + ∞ that this leads to a contradiction.Thus there is an integer k > η for which M − ǫ < h k ( e M ) ≤ M Λ k − Λ k < log e M − M λ k Λ k . Having fixed k this way, we remark that h k ( µ ) is continuous at µ = e M and the lemma follows. (cid:3) Lemma 5.2.
There exist numbers β , e M − < β < e M , and c > , < δ < such that for all µ satisfying β < µ ≤ e M , and for all k satisfying ≤ k ≤ N µ ( N µ is the breakdown index) we have (5.2) e h k ( µ ) µ − h k ( µ ) + M − > c (cid:16) Λ k λ k (cid:17) − δ . Proof.
We apply Lemma 5.1 with η large enough so that the following inequality holds for anyinteger k ≥ η :(5.3) λ k Λ k +1 (cid:12)(cid:12)(cid:12) − Λ k λ k log (cid:16) λ k +1 λ k (cid:17) + 1 − M (cid:12)(cid:12)(cid:12) + Λ k Λ k +1 + ∞ X i =2 i (cid:16) λ k Λ k (cid:17) i < λ k Λ k Λ k +1 . We shall also choose ǫ small enough so that we obtain values of k and β . Without loss of generality,we may assume µ < e M and for the time being we keep µ fixed ( β < µ < e M ) and we write h k instead of h k ( µ ). PENG GAO
As we remarked in Section 3, the sequence h k , h k +1 , . . . is increasing, possibly until breakdown.We shall now first consider those integers k ≥ k for which h k < log µ . For those k we can prove(5.4) h k +1 − h k < λ k Λ k +1 (cid:16)
12 (log µ − h k ) + log( e M µ ) + 34 λ k Λ k (cid:17) . This follows by (3.3) and (3.4), using e − u < − u + u /
2, where u = log µ − h µ and noting that λ k Λ k +1 (cid:16) − Λ k λ k log (cid:16) λ k +1 λ k (cid:17) + 1 − M (cid:17) + Λ k Λ k +1 + ∞ X i =2 i (cid:16) λ k Λ k e − u (cid:17) i < λ k Λ k +1 (cid:12)(cid:12)(cid:12) − Λ k λ k log (cid:16) λ k +1 λ k (cid:17) + 1 − M (cid:12)(cid:12)(cid:12) + Λ k Λ k +1 + ∞ X i =2 i (cid:16) λ k Λ k (cid:17) i < λ k Λ k Λ k +1 , because of e − u < µ < e M and by Lemma 5.1, M − ǫ < h k ≤ h k < log µ , we have 0 < log µ − h k < ǫ , andtherefore we can replace (5.4) by the linear recurrence relation(5.5) h k +1 − h k < λ k Λ k +1 (cid:16) ǫ (log µ − h k ) + log( e M µ ) + 34 λ k Λ k (cid:17) . Putting(5.6) ǫ (log µ − h k ) + log( e M µ ) − λ k Λ k = t k , so that it follows from (5.5) that t k +1 > t k (cid:16) − ǫλ k Λ k +1 (cid:17) + ( λ k k − λ k +1 k +1 − ǫλ k Λ k Λ k +1 ) . As we have assumed that { λ k } ∞ k =1 a non-decreasing sequence, we have λ k k − λ k +1 k +1 − ǫλ k Λ k Λ k +1 ≥ λ k k − λ k +1 k +1 − ǫλ k λ k +1 Λ k Λ k +1 . It follows from (1.8) that the right-hand side expression above is positive if we choose ǫ smallenough and we may assume that our 0 < ǫ < / < log µ − h k < ǫ <
1. It follows that(5.7) t k +1 > t k (cid:16) − ǫλ k Λ k +1 (cid:17) ≥ t k (cid:16) − ǫλ k +1 Λ k +1 (cid:17) . By Lemma 5.1 we have t k > t k > k under consideration.It follows from (5.7) and 1 − ǫx > (1 − x ) ǫ , < x < t k +1 > t k (Λ k ) ǫ (Λ k +1 ) − ǫ = t k ( Λ k λ k ) ǫ ( Λ k +1 λ k ) − ǫ . It follows from (1.8) that the sequence { Λ k /λ k } ∞ k =1 is increasing and we deduce that(5.8) t k +1 > t k ( Λ k λ k ) ǫ ( Λ k +1 λ k ) − ǫ , for all k under consideration, i.e. for all k for which h k < log µ . This is certainly satisfied if t k > log( e M /µ ), and (5.8) guarantees that this is true as long as the right-hand side expression of(5.8) is > log( e M /µ ). Therefore(5.9) t k ≥ t k ( Λ k λ k ) ǫ ( Λ k λ k − ) − ǫ INITE SECTIONS OF WEIGHTED CARLEMAN’S INEQUALITY 9 for all k ≥ k satisfying(5.10) Λ k λ k − < (cid:16) Λ k λ k (cid:17) t /ǫk (cid:16) log( e M /µ ) (cid:17) − /ǫ , and we are sure that no breakdown occurs in this range.Now we return to the discussion on (5.2) and if 0 < h < log µ , we have, on using e − u > − u + u / < u < < log( e M /µ ) <
1, that e h − log µ − h + M − > log( e M /µ ) + 13 (log µ − h ) > (cid:16) log( e M /µ ) (cid:17) + 13 (log µ − h ) > (cid:16) e M /µ ) + log µ − h (cid:17) , where the last inequality above follows from u + v / > u + ( v/ ≥ ( u + v/ / u, v > h = h k and note that it follows from (5.6) and (5.9) that( ǫ + 1) (cid:16) log µ − h k + 2 log( e M µ ) (cid:17) > t k ≥ t k ( Λ k λ k ) ǫ ( Λ k λ k − ) − ǫ , This implies that the left-hand side of (5.2) is at least t k ǫ + 1) ( Λ k λ k ) ǫ ( Λ k λ k − ) − ǫ . This holds for k when (5.10) is satisfied. It follows from (1.4) that λ k /λ k − is bounded above forany k ≥
2. Let c denote such an upper bound and we conclude that the left-hand side of (5.2) isat least t k ǫ + 1) c ǫ ( Λ k λ k ) ǫ ( Λ k λ k ) − ǫ := c ( Λ k λ k ) − ǫ . Other k ’s do not cause much trouble. First, for the values 1 ≤ k < k , we have h k ( µ ) ≤ h k ( µ ) < log µ − M λ k / (2Λ k ) by Lemma 5.1 and the fact that h k increases as k increases. It follows that e h k − log µ − h k + M − >
13 (log µ − h k ) > M λ k k ≥ M λ k k ( Λ k λ k ) − ǫ := c ( Λ k λ k ) − ǫ . Now, for the remaining case k ≤ k ≤ N µ (which is empty if µ = e M ) such thatΛ k λ k − ≥ (cid:16) Λ k λ k (cid:17) t /ǫk (cid:16) log( e M /µ ) (cid:17) − /ǫ , we use that e h − log µ − h + M − > log( e M /µ )for all h to see that the left-hand side of (5.2) is at least (cid:16) Λ k λ k (cid:17) ǫ t k c ǫ ( Λ k λ k ) − ǫ := c ( Λ k λ k ) − ǫ . In all three cases the constants are independent of µ and k , so on letting c = min( c , c , c ) and δ = 2 ǫ completes the proof of the lemma. (cid:3) Lemma 5.3.
There exist numbers β , e M − < β < e M such that for all µ satisfying β < µ < e M there exists an index N < N µ with h N > max(2 , M ) . Proof.
We apply Lemma 5.1 with η large enough and some 0 < ǫ <
1, so that the followingestimation holds for any integer k ≥ η :(5.11) λ k / Λ k < e − − max(2 , M ) / , (cid:12)(cid:12)(cid:12) − Λ k λ k log (cid:16) λ k +1 λ k (cid:17) + 1 − M (cid:12)(cid:12)(cid:12) < − log 2 , and Lemma 5.1 provides us with k > η and β such that (5.1) holds. We now consider the numbers h k , h k +1 , . . . as far as they are < max(2 , M ) + M + 1. If k ≥ k , h k < max(2 , M ) + M + 1, wehave(5.12) µ − λ k / Λ k e h k ≤ / , so that by our definition of the breakdown index (see (3.1)), we have k < N µ . It also follows from(3.3)-(3.5), on using e h /µ − h + M − > log( e M /µ ), that h k +1 − h k > λ k Λ k +1 log( e M /µ ) + O ( λ k Λ k Λ k +1 ) . The lower bound above shows that not for all k ≥ k we have h k ≤ max(2 , M ), since P + ∞ k ( h k +1 − h k ) would diverge in view of (1.7).Now, (5.12), implies that (with u = log µ − h µ here)Λ k Λ k +1 + ∞ X i =2 i (cid:16) λ k Λ k e − u (cid:17) i ≤ + ∞ X i =2 i ( 12 ) i = log 2 − / . It follows from this and (3.3), (3.4), (5.11), (5.12) that h k +1 − h k < λ k Λ k +1 ( e h k µ − µ + M −
1) + 1 − log 2 + log 2 − / . When M ≤
1, the above can be estimated by, via (5.11), h k +1 − h k < λ k Λ k e h k µ + 1 / < < M + 1 . Similarly, when
M >
1, we get h k +1 − h k < λ k Λ k e h k µ + M − / < M + 1 . It follows from the above that if we let h k be the last one below max(2 , M ), then h k +1 is stillbelow max(2 , M ) + M + 1 so that we can take N = k + 1 here and this completes the proof. (cid:3) Proof of Theorem 1.2
As suggested by the discussion in Section 4, we shall study θ ( h k ), where θ is defined by θ ( y ) = Z y dxe x /µ − x + M − . We first simplify the recurrence formula (3.3). Assuming(6.1) e M − < µ ≤ e M , h k < max(2 , M ) , we may also assume k is large enough so that (3.1) is not satisfied. We have h k +1 − h k = λ k Λ k +1 (cid:16) e h k µ − h k + M − γ k (cid:17) , where | γ k | ≤ (cid:12)(cid:12)(cid:12) − Λ k λ k log (cid:16) λ k +1 λ k (cid:17) − M + 1 (cid:12)(cid:12)(cid:12) + Λ k λ k + ∞ X i =2 i (cid:16) λ k Λ k e h µ − log µ (cid:17) i ≤ C λ k Λ k , INITE SECTIONS OF WEIGHTED CARLEMAN’S INEQUALITY 11 for some constant C >
0. It follows from this that there exists a constant C > | h k +1 − h k | ≤ C λ k Λ k +1 . We then deduce easily from above that for h k ≤ x ≤ h k +1 , (cid:12)(cid:12)(cid:12) e x µ − x − ( e h k µ − h k ) (cid:12)(cid:12)(cid:12) ≤ C λ k Λ k +1 ≤ C λ k Λ k , where C > µ or k (still assuming (6.1)).We now apply the mean value theorem to get: θ ( h k +1 ) − θ ( h k ) = ( h k +1 − h k ) θ ′ ( x )with some x in between h k and h k +1 . Hence it follows from our discussion above that θ ( h k +1 ) − θ ( h k ) = λ k Λ k +1 H + γ k H + γ ′ k , where H = e h k µ − h k + M − , | γ k | ≤ C λ k Λ k , | γ ′ k | ≤ C λ k Λ k . We now apply Lemma 5.2 to conclude that there exists a β with e M − < β < e M and a c > < δ < k satisfying 1 ≤ k ≤ N µ , we have H > c (cid:16) Λ k λ k (cid:17) − δ . This implies that | γ k | ≤ C c (cid:16) λ k Λ k (cid:17) − δ H, | γ ′ k | ≤ C c (cid:16) λ k Λ k (cid:17) − δ H. Note it follows from (1.7) that λ k Λ k +1 − λ k Λ k = − λ k λ k +1 Λ k Λ k +1 = O ( 1 k ) . It follows from this and (1.7) that we can find an integer m , independent of µ such that for k > m, h k < max(2 , M ), we have θ ( h k +1 ) − θ ( h k ) = λ k Λ k H + γ k H + γ ′ k + O ( 1 k ) = Ck + O (cid:16) k + 1 k − δ (cid:17) . We recast the above as | θ ( h k +1 ) − θ ( h k ) − C log(1 + 1 /k ) | = O (cid:16) k + 1 k − δ (cid:17) . Now assuming µ < e M , we take the sum over the values m ≤ k < N , where N is the first indexwith h N > max(2 , M ) (see Lemma 5.3). This gives us | θ ( h N ) − C log N | = O (1) + log m + θ ( h m ) . By Lemma 5.1, for any η > M , there exists β , β < β < e M and k > η so that h k ( µ ) < log µ − M λ k / (2Λ k ). We now further take the integer m to be equal to this k . Thus, the maximumof the integrand in θ ( h m ) is attained at x = h m and that e h m /µ − h m + M − ≥ e − Mλ m / (2Λ m ) − log µ + M λ m / (2Λ m ) + M − > − M λ m / (2Λ m ) + M λ m / (8Λ m ) − log µ + M λ m / (2Λ m ) + M − M − log µ + M λ m / (8Λ m ) > M λ m / (8Λ m ) . It follows that θ ( h m ) = Z h m dxe x /µ − x + M − < (log µ )(8Λ m ) / ( M λ m ) = O (1) . We deduce from this that(6.2) | θ ( h N ) − C log N | = O (1) . It is not difficult to find the asymptotic behavior of θ ( ∞ ). If µ < e M , µ → e M , then routinemethods (cf. Sec. 4) lead to θ ( ∞ ) = Z ∞ dxe x /µ − x + M − √ π (cid:16) log( e M /µ ) (cid:17) − / + O (1) . It is also easy to see that θ ( ∞ ) − θ (max(2 , M )) = O (1). As h N ≥ max(2 , M ), we have θ (max(2 , M )) ≤ θ ( h N ) < θ ( ∞ ). It follows from (6.2) thatlog N = √ πC (cid:16) log( e M /µ ) (cid:17) − / + O (1) . According to (3.8) and our discussion in Section 4, this completes the proof of (4.2) and it wasalready shown there that (4.2) leads to our assertion for Theorem 1.2.7.
An Application of Theorem 1.2
As an application of Theorem 1.2, we consider in this section the case λ k = k α for α ≥ { k α } ∞ k =1 is a non-decreasing sequence satisfying (1.4). We note the following Lemma 7.1.
Let α ≥ be fixed. For any integer n ≥ , we have (7.1) αα + 1 n α ( n + 1) α ( n + 1) α − n α ≤ n X i =1 i α ≤ ( n + 1) α +1 α + 1 . We point out here the left-hand side inequality above is [6, Lemma 2, p.18] and the right-handside inequality can be easily shown by induction.It follows readily from the above lemma that (1.7) holds with C = α + 1. We note here it is easyto see that (1.2) with M = 1 /C follows from the left-hand side inequality of (7.1), which implies n +1 X i =1 i α / ( n + 1) α − n X i =1 i α /n α = 1 + (cid:16) n + 1) α − n α (cid:17) n X i =1 i α ≤ α + 1 . This combined with the upper bound in (7.1) also leads to (1.6) easily.Now, to show (1.5), we assume (1.8) for the moment and note thatlog (cid:16) Λ k +1 /λ k +1 Λ k /λ k (cid:17) = log (cid:16) k +1 /λ k +1 − Λ k /λ k Λ k /λ k (cid:17) ≥ Λ k +1 /λ k +1 − Λ k /λ k Λ k +1 /λ k +1 . We then deduce that (1.5) follows fromΛ k +1 /λ k +1 − Λ k /λ k ≥ M λ k /λ k +1 . Note that the above also establishes (1.8). In our case, it is easy to see that this becomes (for any n ≥ n X i =1 i α ≤ αα + 1 ( n + 1) α ( n + 2) α − ( n + 1) α . To show this, we define P n ( α ) = n n X i =1 i α (cid:30) n + 1 n +1 X i =1 i α ! /α . INITE SECTIONS OF WEIGHTED CARLEMAN’S INEQUALITY 13
We recall that Bennett [1] proved that for α ≥ P n ( α ) ≤ P n (1) = n + 1 n + 2 . It is easy to see that this is equivalent to n X i =1 i α ≤ n ( n + 1) α ( n + 1)( n + 2) α − n ( n + 1) α . Thus, in order to prove (7.2), it suffices to prove the following n ( n + 1) α ( n + 1)( n + 2) α − n ( n + 1) α ≤ αα + 1 ( n + 1) α ( n + 2) α − ( n + 1) α . The above inequality can be seen easily to be equivalent to the following n (cid:16) ( n + 2) α − ( n + 1) α (cid:17) ≤ α ( n + 2) α , which follows easily from the mean value theorem. Thus, we have shown, as a consequence ofTheorem (1.2) the following Corollary 7.1.
Fix α ≥ and let λ k = k α for k ≥ . Then inequality (1.3) holds with U N = e / ( α +1) − π e / ( α +1) ( α + 1) (log N ) + O (cid:16) N ) (cid:17) . References [1] G. Bennett, Lower bounds for matrices. II.,
Canad. J. Math. , (1992), 54-74.[2] T. Carleman, Sur les fonctions quasi-analytiques, in Proc. 5th Scand. Math. Congress , Helsingfors, Finland, 1923,pp. 181–196.[3] N. G. De Bruijn, Carleman’s inequality for finite series,
Nederl. Akad. Wetensch. Proc. Ser. A 66 = Indag,Math. , (1963), 505–514.[4] P. Gao, A note on Carleman’s inequality, arXiv:0706.2368.[5] J. Duncan and C. M. McGregor, Carleman’s inequality, Amer. Math. Monthly , (2003), 424–431.[6] V. I. Levin and S.B. Steˇckin, Inequalities, Amer. Math. Soc. Transl. (2) , (1960), 1–29.[7] J. Peˇcari´c and K. Stolarsky, Carleman’s inequality: history and new generalizations, Aequationes Math. , (2001), 49–62. Department of Computer and Mathematical Sciences, University of Toronto at Scarborough, 1265Military Trail, Toronto Ontario, Canada M1C 1A4
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