On Copson's inequalities for $0

Let $(\lambda_n)_{n \geq 1}$ be a non-negative sequence with $\lambda_1>0$ and let $\Lambda_n=\sum^n_{i=1}\lambda_i$. We study the following Copson inequality for $0p$, \begin{align*} \sum^{\infty}_{n=1}\left (\frac 1{\Lambda_n} \sum^{\infty}_{k=n}\lambda_k x_k \right )^p \geq \left ( \frac {p}{L-p}\right )^p \sum^{\infty}_{n=1}x^p_n. \end{align*} We find conditions on $\lambda_n$ such that the above inequality is valid with the constant being best possible.


Introduction
Let p > 0 and x = (x n ) n≥1 be a positive sequence Let (λ n ) n≥1 be a positive sequence and let The above two inequalities are equivalent (see [10]) and the constants are best possible. When λ k = 1, k ≥ 1 and c = p, inequality (1.1) becomes the following celebrated Hardy inequality ([12, Theorem 326]): It is noted in [12] that the constant p p in (1.4) may not be best possible and the best constant for 0 < p ≤ 1/3 was shown by Levin and Stečkin [13, Theorem 61] to be indeed (p/(1p)) p . In [8], it is shown that the constant (p/(1p)) p stays best possible for all 0 < p ≤ 0.346. It is further shown in [11] that the constant (p/(1p)) p is best possible when p = 0.35.
There exists an extensive and rich literature on extensions and generalizations of Copson's inequalities and Hardy's inequality (1.3) for p > 1. For recent developments in this direction, we refer the reader to the articles in [6][7][8][9][10][11] and the references therein. On the contrary, the case 0 < p < 1 is less known as can be seen by comparing inequalities (1.3) and (1.4). On one hand, the constant in (1.3) is shown to be best possible for all p > 1. On the other hand, though it is known the best constant that makes inequality (1.4) valid is (p/(1p)) p when 0 < p ≤ 0.346, it is shown in [8] that the constant (p/(1p)) p fails to be best possible when 1/2 ≤ p < 1 and the best constant in these cases remains unknown.
Our goal in this paper is to study the following variation of Copson's inequalities for 0 < p < 1: It is an open problem to determine the best possible constant to make inequality (1.5) valid in general. Our choice for presenting the constant in the form (p/(Lp)) p in (1.5) is motivated by the study on the analogue case of inequality (1.5) when p > 1. We define A result of Cartlidge [3] shows that when L λ < p for p > 1, then the following inequality holds for all non-negative sequences x: We shall see in Theorem 1.3 below that the constant given in inequality (1.5) is indeed best possible for certain sequences (λ n ) and certain ranges of p when one replaces L by L λ in (1.5). This includes case concerning the classical inequality (1.4) (with p p replaced by (p/(1p)) p there).
Further, let q < 0 be the number satisfying 1/p + 1/q = 1, we note that inequality (1.5) is equivalent to its dual version (assuming that x n > 0 for all n): The equivalence of the above two inequalities can be easily established following the discussions in [8,Sect. 1].
Our main result gives a condition on λ n and L such that inequalities (1.5) and (1.8) hold. For this purpose, we define, for constants p and L, (1.9) Our main result is the following statement.
Note that the values of p are not given explicitly in (1.10), nor by the conditions a i (L λ , p) ≥ 0, i = 1, 2. Thus, Theorem 1.1 is not readily applied in practice. For this reason, and with future applications in mind, we develop the following result.
then inequality (1.5) holds with L replaced by L λ there for all non-negative sequences x.
Suppose that there exist positive constants 1/2 < L < 1, 0 < M < 1, L + 2M < 1 such that, for any integer n ≥ 1, (1.14) Then inequality (1.5) holds for all non-negative sequences x when We remark here that it is easy to see that the minimum on the right-hand side of (1.15) can take either values. We now apply Theorem 1.2 to study inequalities (1.11)-(1.12). As the case α = 1 yields the classical inequality (1.4), we concentrate on the case α > 1 and we deduce readily from Theorem 1.2 the following result. Theorem 1.3 Let α ≥ 1 and p 1/α be defined as in (1.13). Then inequality (1.11) holds for all non-negative sequences x when α > 1, 0 < p ≤ p 1/α and inequality (1.12) holds for all non-negative sequences x when α ≥ 2, 0 < p ≤ p 1/α . The constants are best possible in both cases.

Proof of Theorem 1.1
For the first assertion of Theorem 1.1, our goal is to find conditions on the λ n such that the following inequality holds for 0 < p < 1, L > p: It suffices to prove the above inequality by replacing the infinite sums by finite sums from n = 1 to N (and k = n to N ) for any integer N ≥ 1. Note that, as in [6, Sect. 3], we have, for any positive sequence w = (w n ), Suppose we can find a sequence w = (w n ) of positive terms, such that, for any integer n ≥ 1, .
Then the desired inequality follows. Now we make a change of variables: w i → λ i w i to recast the above inequality as We now define our sequence w = (w n ) to satisfy w 1 = 1 and we inductively see that, for n ≥ 2, This allows us to deduce that, for n ≥ 1, We now prove the second assertion of Theorem 1.1. We set x = λ n /Λ n , y = λ n+1 /Λ n+1 to recast inequality (1.10) as x 1/(1-p) y p/ (1-p) .
To facilitate also the proof of Theorem 1.2 below, we proceed by taking the condition (1.14) into consideration to assume that L is a constant such that 1/y ≤ 1/x + L + Mx, where M ≥ 0 is another constant. As the function t → (t -1) (1+p)/(1-p) t -p/(1-p) is an increasing function of t ≥ 1, we deduce that it suffices to prove the above inequality for 1/y = 1/x + L + Mx, which is equivalent to showing that f L,M,p (x) ≥ 0 for 0 < x ≤ 1, where Suppose that (1.6) is valid and L λ ≥ 1. In this case we set L = L λ and M = 0 so that it suffices to show that f L,0,p (x) ≥ 0. Calculation shows that Suppose that 0 < p ≤ 1/3. We want to show that g L,p (x) ≥ 0 for 0 < x ≤ 1. We first note that we have We may now assume that Otherwise, we have trivially g L,p (x) ≥ 0. We then estimate (1 + (L -1)x) It then follows that g L,p (x) ≥ u L,p (x), where Suppose first that We then deduce that We regard the last expression above as a linear function of x to see that its derivative with respect to x is As we have 0 < p ≤ 1/3 and L ≥ 1, we deduce that It follows from this that the last expression in (2.5) is minimized at x = 0 with corresponding value being u L,p (0). On the other hand, if the inequality in (2.4) does not hold, then one checks that u L,p (x) is a quadratic function of x with negative leading coefficient, hence is minimized at x = 0 or x = 1. Thus, in either case, we conclude that u L,p (x) ≥ min{u L,p (0), u L,p (1)} for 0 ≤ x ≤ 1. One checks that When we regard the above expression as a function of L, it is readily seen that u L,p (0) is convex in L such that We thus deduce that u L,p (0) ≥ 0 when 0 < p ≤ 1/3. On the other hand, we have u L,p (1) = a 1 (L, p) ≥ 0 by our assumption, where a 1 (L, p) is defined in (1.9). It follows that g L,p (x) ≥ u L,p (x) ≥ 0, hence f L,0,p (x) ≥ 0 for 0 < x ≤ 1. As f L,0,p (0) = f L,0,p (0) = 0, we then deduce that f L,0,p (x) ≥ 0 and this completes the proof for the case L λ ≥ 1 of the second assertion of Theorem 1.1.
To prove the case 0 < L λ < 1 of the second assertion of Theorem 1.1, we first note that One checks that It follows from this and the arithmetic-geometric mean inequality that Suppose that (1.6) is valid and 0 < L λ < 1. In this case we can also set L = L λ and M = 0 to see that inequality (2.7) becomes Calculation shows that One checks that v L,p (x) is a quadratic polynomial of x with a negative leading coefficient when L ≥ 2p. It follows that v L,p (x) ≥ min{v L,p (0), v L,p (1)} for 0 < x ≤ 1 and one checks that v L,p (0) = pu L,p (0), where u L,p (0) is defined in (2.6). Similar to our discussions for the case L > 1, one checks that u L,p (0) is convex in L such that

Proof of Theorem 1.2
First, we assume that (1.6) is valid and we set L = L λ in this case. It suffices to find a value of p such that a 2 (L, p) ≥ 0 by Theorem 1.1. Note that lim p→0 + a 2 (ap, p)/p < 0 when a > 1, it is therefore not possible to show a 2 (L, p) ≥ 0 by assuming that p ≤ L/a for any a > 1. We therefore seek to show a 2 (L, p) ≥ 0 for p ≤ L 2 /4. We first note that where the last inequality follows from the observation that the function p → 4 -5p -2 1-p is non-negative for 0 < p ≤ 1/4. This completes the proof for the first assertion of Theorem 1.2.
To prove the second assertion of Theorem 1.2, we see from the proof of Theorem 1.1 that it suffices to show the right-hand side expression of (2.7) is ≥ 1 for 0 < x ≤ 1. We simplify it to see that it is equivalent to showing that We assume that This implies that the function is a concave function of x, hence is minimized at x = 0 or 1. When x = 0, the above function takes the value 1. We further assume that the above function takes a value ≥ 1 when x = 1.
That is, We then deduce that We apply the above estimation and the estimation that 1 + 2Mx/L ≥ 1 in (3.1) to see that it suffices to show that, for 0 < x ≤ 1, We now assume that It is easy to see that v L,M,p (x) is a quadratic polynomial of x with negative leading coefficient The above inequality is certainly valid when L/p = 2. We may thus assume that L/p > 2 to see that the above inequality is a consequence of the following inequality: We apply the above estimates to see that inequality (3.5) is a consequence of the following inequality: L -2p -1p 2 (1 -L) L p ≥ (1 + p)(1 + L + M). (3.8) As inequality (3.7) implies inequalities (3.6) and (3.8), we first find values of p so that inequality (3.7) holds. To do so, we first simply inequality (3.7) by noting that L -2p -1p 2 (1 -L) ≥ 2L -1 -2p.