Inequalities Related to Hardy’s Inequality Trong quá trình cố gắng để đơn giản hóa giấy tờ chứng minh định lý loạt tăng gấp đôi Hilbert, GH Hardy [136] lần đầu tiên chứng minh vào năm 1920 bất bình đẳng nổi tiếng nhất mà bây giờ được biết đến trong các tài liệu như bất bình đẳng của Hardy. Hardy của sự bất bình đẳng đáng kể trong các điều khoản của sự đơn giản của nó, số lượng lớn các kết quả mà nó đề, và sự đa dạng của các ứng dụng có thể được liên quan đến nó | Chapter 2 Inequalities Related to Hardy s Inequality Introduction In the course of attempts to simplify the proof of Hilbert s double series theorem . Hardy 136 first proved in 1920 the most famous inequality which is now known in the literature as Hardy s inequality. Hardy s inequality is remarkable in terms of its simplicity the large number of results to which it deals and the variety of applications which can be related to it. Since from its discovery Hardy s inequality has evoked the interest of many mathematicians and large number of papers have appeared which deal with new proofs various extensions refinements generalizations and series analogues. In the past few years various investigators have discovered many useful and new inequalities related to well-known Hardy s inequality. This chapter presents a number of new and basic inequalities related to Hardy s inequality recently investigated in order to achieve a diversity of desired goals. Hardy s Series Inequality and Its Generalizations There is a vast and growing literature related to the series inequalities. In this section we will give some basic inequalities involving series of terms which find important applications in analysis. In an attempt to give a simple proof of Hilbert s inequality Hardy 136 see also 141 Theorem 315 establishes the following most fundamental inequality. Theorem . If p 1 an f 0 and An a1 a2 - an then 113 114 Chapter 2. Inequalities Related to Hardy s Inequality unless all the a are zero. The constant is the best possible. Proof. The proof given here is due to Elliott 105 and is also given in 141 . By relabeling if necessary we may assume that a1 0 and hence that each An 0. We write an for An n and agree that any number with suffix 0 is equal to 0. Now by making use of the elementary inequality xn 1 nyn 1 n 1 xyn x y 0 reals we observe that p p p 1 p p I an D 1an an an 1 na n n 1 an 1 an z p np an 1 1 p 1 n 1 p p 1 1 . an an 1 p 1 _p A np c an 1 p 1 1
