On an inequality suggested by Littlewood

We study an inequality suggested by Littlewood, our result refines a result of Bennett.


Introduction
In connection with work on the general theory of orthogonal series, Littlewood [7] raised some problems concerning elementary inequalities for infinite series. One of them asks to decide whether an absolute constant K exists such that for any non-negative sequence (a n ) with A n = n k=1 a k , The above problem was solved by Bennett [3], who proved the following more general result: Theorem 4]). Let p ≥ 1, q > 0, r > 0 satisfying (p(q + r) − q)/p ≥ 1 be fixed. Let K(p, q, r) be the best possible constant such that for any non-negative sequence (a n ) with A n = n k=1 a k , (a p n A q n ) 1+r/q .
The special case p = 1, q = r = 2 in (1.2) leads to inequality (1.1) with K = 4 and Theorem 1.1 implies that K(p, q, r) is finite for any p ≥ 1, q > 0, r > 0 satisfying (p(q + r) − q)/p ≥ 1, a fact we shall use implicitly throughout this paper. We note that Bennett only proved Theorem 1.1 for p, q, r ≥ 1 but as was pointed out in [1], Bennett's proof actually works for the p, q, r's satisfying the condition in Theorem 1.1. Another proof of inequality (1.2) for the special case r = q was provided by Bennett in [2] and a close look at the proof there shows that it in fact can be used to establish Theorem 1.1.
On setting p = 2 and q = r = 1 in (1.2), and interchanging the order of summation on the left-hand side of (1.2), we deduce the following The constant in (1.3) was improved to be 2 1/3 in [6] and the following more general result was given in [4]: Then for any non-negative sequence (a n ) with A n = n k=1 a k , Note that inequality (1.3) with constant 2 1/3 corresponds to the case p = 3, q = 2, r = 1 in (1.4). In [8], an even better constant was obtained but the proof there is incorrect. In [1], [4] and [8], results were also obtained concerning inequality (1.2) under the extra assumption that the sequence (a n ) is non-decreasing.
We point out here that among the three expressions on the right-hand side of (1.8), each one is likely to be the minimum. For example, the middle one becomes the minimum when p = 2, q = 1 while it's easy to see that the last one becomes the minimum for p = q large enough and the first one becomes the minimum when q is being fixed and p → ∞. Moreover, it can happen that the minimum value in (1.5) occurs at a δ other than q(p − 1)/(p(q + 1) − q), 1. For example, when p = q = 6, the bound (1.8) gives K(6, 6, 1) ≤ 21/5 while one checks easily that C(6, 6, 1.15/1.2) < 21/5. We shall not worry about determining the precise minimum of (1.5) in this paper.
We note that the special case p = 1, q = r = 2 of Theorem 1.3 leads to the following improvement on Bennet's result on the constant K of inequality (1.1):

A few Lemmas
The constant is best possible.
Lemma 2.2. Let p < 0. For any non-negative sequence (a n ) with a 1 > 0 and A n = n k=1 a k , we have for any n ≥ 1, Proof. We start with the inequality Replacing A k−1 in the middle term of the left-hand side expression above by A k −a k and simplifying, we obtain Inequality (2.1) follows from above upon noting that a n A p−1 n ≤ A p n .
Proof. As it is easy to check the assertion of the lemma holds when p = 1 or r = 1, we may assume p > 1, r > 1 here. We set Note that we have 0 < α < 1 as p > 1, r > 1 here. By Hölder's inequality, we have The assertion of the lemma now follows on applying inequality (1.2) to both factors of the last expression above.
Lemma 2.4. Let p ≥ 1, q > 0, 0 < r ≤ 1 be fixed satisfying (p(q + r) − q)/p ≥ 1. Under the same notions of Theorem 1.1, we have Proof. We may assume 0 < r < 1 here. We set Note that we have 0 < α < 1. By Hölder's inequality, we have The assertion of the lemma now follows on applying inequality (1.2) to the second factor of the last expression above.

Proof of Theorem 1.3
We obtain the proof of Theorem 1.3 via the following two lemmas: Lemma 3.1. Let p ≥ 1, q > 0 be fixed. Under the same notions of Theorem 1.1, inequality (1.2) holds when r = 1 with K(p, q, 1) bounded by the right-hand side expression of (1.5).
Proof. We may assume that only finitely many a n 's are positive, say a n = 0 whenever n > N . We may also assume a 1 > 0. As the case p = 1 of the lemma is already contained in Theorem 1.1, we may further assume p > 1 throughout the proof. Moreover, even though the assertion that K(p, q, 1) ≤ (p(q + 1) − q)/p is already given in Theorem 1.1, we include a new proof here.

Now, we set
so that P, Q > 1 and 1/P + 1/Q = 1. We then have, by Hölder's inequality, It follows from inequalities (3.4) and (3.5 One sees easily that the above inequality also holds when δ = q(p − 1)/(p(q + 1) − q). Combining the above inequality with (3.3), we see this establishes (3.2) with where C(p, q, δ) is defined as in (1.6) and the minimum is taken over the δ's satisfying (1.7) and this completes the proof of Lemma 3.1.
Lemma 3.2. Let p = 1, q > 0, r ≥ 1 be fixed. Under the same notions of Theorem 1.1, we have Proof. We may assume a n = 0 whenever n > N . In this case, on setting b n = a n A q n , c n = Note that as r ≥ 1, we have the following bounds: We then apply partial summation together with the bounds above to obtain (with When r ≥ 2, we apply inequality (2.2) to see that Combining this with inequality (3.6), we see that this implies that The assertion of the lemma for r ≥ 2 now follows on applying inequality (1.2) to the right-hand side expression above. When 1 ≤ r ≤ 2, we apply inequality (2.3) in (3.6) to see that .
The bound for K(p, q, 1) follows from Lemma 3.1 and this completes the proof of Theorem 1.3.

Further Discussions
We now look at inequality (1.2) in a different way. For this, we define for any non-negative sequence (a n ) and any integers N ≥ n ≥ 1, We then note that in order to establish inequality (1.2), it suffices to show that for any integer N ≥ 1, we have (a p n A q n ) 1+r/q .
Upon a change of variables: a n → a N −n+1 and recasting, we see that the above inequality is equivalent to Here K(p, q, r) is also the best possible constant such that inequality (4.1) holds for any non-negative sequence (a n ).
We point out that one can give another proof of Theorem 1.3 by studying (4.1) directly. As the general case r ≥ 1 can be reduced to the case r = 1 in a similar way as was done in the proof of Theorem 1.3 in Section 3, one only needs to establish the upper bound for K(p, q, 1) given in (1.5). For this, one can use an approach similar to that taken in Section 3, in replacing Lemma 2.1 and Lemma 2.2 by the following lemmas. Due to the similarity, we shall leave the details to the reader. Lemma 4.1. Let d ≥ c > 1 and (λ n ) be a positive sequence with ∞ k=1 λ k < ∞. Let Λ * n = ∞ k=n λ k . Then for all non-negative sequences (x n ), The constant is best possible.
The above lemma is Corollary 6 to Theorem 2 of [3] and only the special case d = c is needed for the proof of Theorem 1.3.