Mathematical Inequalities - Chapter 2 ppt

Chapter 2 Inequalities Related to Hardy’s Inequality 2.1 Introduction In the course of attempts to simplify the proof of Hilbert’s double series theo- rem, G.H. 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 re- markable 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 dis- covery Hardy’s inequality has evoked the interest of many mathematicians, and large number of papers have appeared which deal with new proofs, various ex- tensions, 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. 2.2 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. T HEOREM 2.2.1. If p>1, a n  0 and A n =a 1 +a 2 +···+a n , then ∞  n=1  A n n  p <  p p −1  p ∞  n=1 a p n , (2.2.1) 113 114 Chapter 2. Inequalities Related to Hardy’s Inequality unless all the a are zero. The constant is the best possible. P ROOF. The proof given here is due to Elliott [105] and is also given in [141]. By relabeling, if necessary, we may assume that a 1 > 0 and hence that each A n > 0. We write α n for A n /n and agree that any number with suffix 0 is equal to 0. Now, by making use of the elementary inequality x n+1 +ny n+1  (n +1)xy n , (2.2.2) x,y  0 reals, we observe that α p n − p p −1 α p−1 n a n = α p n − p p −1  nα n −(n −1)α n−1  α p−1 n = α p n  1 − np p −1  + (n −1)p p −1 α p−1 n α n−1  α p n  1 − np p −1  + n −1 p −1  (p −1)α p n +α p n−1  = 1 p −1  (n −1)α p n−1 −nα p n  . (2.2.3) By substituting n =1, 2, ,N in (2.2.3) and adding the inequalities we have N  n=1 α p n − p p −1 N  n=1 α p−1 n a n  − Nα p N p −1  0. (2.2.4) From (2.2.4) we observe that N  n=1 α p n  p p −1 N  n=1 α p−1 n a n . (2.2.5) Using Hölder’s inequality with indices p, p/(p − 1) on the right-hand side of (2.2.5) we have N  n=1 α p n  p p −1  N  n=1 a p n  1/p  N  n=1 α p n  (p−1)/p . (2.2.6) Dividing by the last factor on the right-hand side (which is certainly positive) and raising the result to the pth power, we get N  n=1 α p n   p p −1  p N  n=1 a p n . (2.2.7) 2.2. Hardy’s Series Inequality and Its Generalizations 115 When we make N tend to infinity we obtain (2.2.1), except that we have “less than or equal to” in place of “less than”. In particular, we see that  ∞ n=1 α p n is finite. Returning to (2.2.5), and replacing N by ∞, we obtain ∞  n=1 α p n  p p −1 ∞  n=1 α p−1 n a n  p p −1  ∞  n=1 a p n  1/p  ∞  n=1 α p n  (p−1)/p . (2.2.8) There is an inequality in the second place unless (a p n ) and (α p n ) are proportion, that is, unless a n =Cα n , where C is independent of n. If this is so then (a 1 =α 1 > 0) C must be 1, and then A n = na n for all n. This idea is inconsistent with the convergence of  ∞ n=1 a p n . Hence ∞  n=1 α p n < p p −1  ∞  n=1 a p n  1/p  ∞  n=1 α p n  (p−1)/p , (2.2.9) and (2.2.1) follows from (2.2.9) as (2.2.7) followed from (2.2.6). To prove the constant factor the best possible, we take a n =n −1/p ,n N, a n =0,n>N. Then ∞  n=1 a p n = N  n=1 1 n , A n = n  1 v −1/p >  n 1 x −1/p dx = p p −1  n (p−1)/p −1  ,n N,  A n n  p >  p p −1  p  1 −ε n n  ,n N, where ε n →0 when n →∞. It follows that ∞  n=1  A n n  p > N  n=1  A n n  p >  p p −1  p (1 −η N ) ∞  n=1 a p n , 116 Chapter 2. Inequalities Related to Hardy’s Inequality where η N →0 when N →∞. Hence, any inequality of the type ∞  n=1  A n n  p <  p p −1  p (1 −ε) ∞  n=1 a p n is false if a n is chosen as above and N is sufficiently large.  The above theorem states the relationship between the arithmetic means of a sequence and the sequence itself. This theorem along with its integral analogue was first proved by Hardy, which later went by the name “Hardy’s inequality”. The constant at the right-hand side of (2.2.1) is determined by Landu in [182], who showed that it is the best possible for each p. There are many generalizations and extensions of Theorem 2.2.1, which have been proved by different writers in different ways; and we give some of these results here in the following theorems. In 1926, Copson [69] generalizes Theorem 2.2.1 by replacing the arithmetic mean of a sequence by a weighted arithmetic mean. We shall first consider the following version of Copson’s generalization of Hardy’s inequality. T HEOREM 2.2.2. Let p>1, λ n > 0, a n > 0, n =1, 2, ,  ∞ n=1 λ n a p n converge, and further let Λ n =  n i=1 λ i , A n =  n i=1 λ i a i . Then ∞  n=1 λ n  A n Λ n  p   p p −1  p ∞  n=1 λ n a p n . (2.2.10) P ROOF. We write α n = A n Λ −1 n and agree that any number with suffix 0 is equal to 0. Now, by making use of the elementary inequality (2.2.2), we observe that λ n α p n − p p −1 λ n a n α p−1 n =λ n α p n − p p −1 α p−1 n [Λ n α n −Λ n−1 α n−1 ] =  λ n −Λ n p p −1  α p n + p p −1 Λ n−1 α n−1 α p−1 n   λ n −Λ n p p −1  α p n + Λ n−1 p −1  α p n−1 +(p −1)α p n  = 1 p −1  (pλ n −λ n −pΛ n +pΛ n−1 −Λ n−1 )α p n +Λ n−1 α p n−1  2.2. Hardy’s Series Inequality and Its Generalizations 117 = 1 p −1  pλ n −λ n −p(Λ n −Λ n−1 ) −Λ n−1  α p n +Λ n−1 α p n−1  = 1 p −1  (pλ n −λ n −pλ n −Λ n−1 )α p n +Λ n−1 α p n−1  = 1 p −1  −Λ n α p n +Λ n−1 α p n−1  = 1 p −1  Λ n−1 α p n−1 −Λ n α p n  . (2.2.11) By substituting n =1, ,N in (2.2.11) and adding the inequalities, we have N  n=1 λ n α p n − p p −1 N  n=1 λ n a n α p−1 n  − 1 p −1 Λ n α p N  0. (2.2.12) From (2.2.12) we observe that N  n=1 λ n α p n  p p −1 N  n=1 λ n a n α p−1 n . (2.2.13) Using Hölder’s inequality with indices p, p/(p − 1) on the right-hand side of (2.2.13) we have N  n=1 λ n α p n  p p −1  N  n=1 λ n a p n  1/p  N  n=1 λ n α p n  (p−1)/p . Dividing the above inequality by the last factor on the right-hand side and raising the result to the pth power, we obtain N  n=1 λ n α p n   p p −1  p N  n=1 λ n a p n . (2.2.14) When we make N tend to infinity we obtain (2.2.10).  A version of the companion inequality proved by Copson [69] can be stated as follows. 118 Chapter 2. Inequalities Related to Hardy’s Inequality THEOREM 2.2.3. Let p>1, λ n > 0, a n > 0 for n = 1, 2, ,  ∞ n=1 λ n a p n con- verge, and further let Λ n = n  i=1 λ i ,A n = ∞  i=n λ i a i Λ i . Then ∞  n=1 λ n A p n  p p ∞  n=1 λ n a p n . (2.2.15) As in the proof of Theorem 2.2.1, see also Copson [69, p. 12], the constants in- volved in (2.2.10) and (2.2.15) are best possible. In 1928, Hardy in his paper [137] notes that the inequality given in Theorem 2.2.3 does not require a separate proof but can be derived from Copson’s first inequality given in Theorem 2.2.2. In view of this remark, here we omit the proof of Theorem 2.2.3. For an independent proof of Theorem 2.2.3, see Copson [69]. In [139] Hardy and Littlewood generalizes Hardy’s inequality in Theo- rem 2.2.1 as follows. T HEOREM 2.2.4. Suppose p>0, c is a real (but not necessarily positive) con- stant and  ∞ n=1 a n is a series of positive terms. Set A 1n = n  k=1 a k and A n∞ = ∞  k=n a k . If p>1 we have ∞  n=1 n −c A p 1n  K ∞  n=1 n −c (na n ) p with c>1, (2.2.16) ∞  n=1 n −c A p n∞  K ∞  n=1 n −c (na n ) p with c<1, (2.2.17) and if p<1 we have ∞  n=1 n −c A p 1n  K ∞  n=1 n −c (na n ) p with c>1, (2.2.18) ∞  n=1 n −c A p n∞  K ∞  n=1 n −c (na n ) p with c<1, (2.2.19) 2.2. Hardy’s Series Inequality and Its Generalizations 119 where K denotes a positive constant, not necessarily the same at each occurrence. Theorem 2.2.4 was generalized by Leindler in [186], who replaced in (2.2.16)–(2.2.19) the sequence {n −c } by an arbitrary sequence {λ n }: for instance, he proved the inequality ∞  n=1 λ n A p 1n  p p ∞  n=1 λ 1−p n  ∞  m=n λ m  p a p n (2.2.20) with p  1 and λ n > 0. In [226] Nemeth gives further generalizations by combing Hardy’s inequality in Theorem 2.2.1 and the Hardy and Littlewood inequality in Theorem 2.2.4. In the following theorem we present the results given in [226]. We use the following definitions given in [226]. (i) C ∈M 1 denotes that the matrix C =(c m,v ) satisfies the conditions: c m,v > 0,v m, c m,v =0, v>m,m,v=1, 2, , and 0 < c m,v c n,v  N 1 , 0  v  n  m. (2.2.21) (ii) C ∈ M 2 denotes that c m,v > 0 (v  m) and c m,v = 0 (v < m,m,v = 1, 2, ), c m,v c n,v  N 2 , 0  n  m  v. (2.2.22) (iii) C ∈ M 3 denotes that c v,m > 0 (v  m) and c v,m = 0 (v < m,v,m = 1, 2, ), 0 < c v,m c v,n  N 3 ,v n  m  0. (2.2.23) (iv) C ∈ M 4 denotes that c v,m > 0 (v  m) and c v,m = 0 (v > m,v,m = 1, 2, ), c v,m c v,n  N 4 , 0  v  m  n, (2.2.24) where N i denote positive absolute constants for i =1, 2, 3, 4. The main result given by Nemeth in [226] follows. T HEOREM 2.2.5. Let a n  0 and λ n > 0, n = 1, 2, , be given, and let C =(c m,k ) be a triangular matrix. 120 Chapter 2. Inequalities Related to Hardy’s Inequality (a) If C ∈M 1 and p  1, then ∞  n=1 λ n  n  m=1 c n,m a m  p  N p(p−1) 1 p p ∞  n=1 λ 1−p n  ∞  m=n λ m c m,n  p a p n . (2.2.25) (b) If C ∈M 3 and p  1, then ∞  m=1 λ m  ∞  n=m c n,m a n  p  N p(p−1) 3 p p ∞  m=1 λ 1−p m  m  n=1 λ n c m,n  p a p m . (2.2.26) (c) If C ∈M 2 and 0 <p 1, then ∞  n=1 λ n  ∞  v=n c n,v a v  p  N (1−p)p 2 p p ∞  n=1 λ 1−p n  n  k=1 c k,n λ k  p a p n . (2.2.27) (d) If C ∈M 4 and 0 <p 1, then ∞  m=1 λ m  m  n=1 c n,m a n  p  N (1−p)p 4 p p ∞  m=1 λ 1−p m  ∞  n=m λ n c m,n  p a p m . (2.2.28) We note that Theorem 2.2.5 implies Leindler’s theorem in [186], further if λ m =c m,m f 1−p (m) and we write c m,n f (m) instead of elements of the matrix C, then assertion (a) includes Theorem 3 of Izumi, Izumi and Petersen [163], and in the case λ n =f −p (n) and c k,n =f(k)a k,n , assertion (d) reduces to Theorem 5 of Davis and Petersen [78]. In the proof of the above theorem, we require the following lemmas. L EMMA 2.2.1 [78, Lemma 1]. If p>1 and z n  0, n = 1, 2, ,then  n  k=1 z k  p  p n  k=1 z k  k  v=1 z v  p−1 . P ROOF.Letλ r =z 1 +···+z r , r = 1, 2, ,n. Then (λ n ) p = p  λ n 0 x p−1 dx = p   λ 1 0 +  λ 2 λ 1 +···+  λ n λ n−1  x p−1 dx 2.2. Hardy’s Series Inequality and Its Generalizations 121  p  λ 1 λ p−1 1 +(λ 2 −λ 1 )λ p−1 2 +···+(λ n −λ n−1 )λ p−1 n  = p  z p 1 +z 2 (z 1 +z 2 ) p−1 +···+z n (z 1 +···+z n ) p−1  . The proof is complete.  The proofs of the following lemmas are similar to that of Lemma 2.2.1 (see [226]). L EMMA 2.2.2. If 0 <p<1 and z 1 > 0, z n  0, n = 2, 3, ,then  n  k=1 z k  p  p n  k=1 z k  k  v=1 z v  p−1 . L EMMA 2.2.3. If 0 <p<1 and z n  0, n = 1, 2, , then for every natural number N , for which z N > 0,  N  k=n z k  p  p N  k=n z k  N  v=k z v  p−1 . L EMMA 2.2.4. If p>1 and z n  0, n = 1, 2, , then for every natural num- ber N ,  N  k=n z k  p  p N  k=n z k  N  v=k z v  p−1 . P ROOF OF THEOREM 2.2.5. For p = 1 the assertions are obvious; we have only to interchange the order of summations. Further we may assume that not all a n vanish (otherwise the theorem is evident). (a) By Lemma 2.2.1 we obtain, for C =(c m,k ) ∈M 1 , N  n=1 λ n  n  m=1 c n,m a m  p  p N  n=1 λ n n  m=1 c n,m a m  m  k=1 c n,k a k  p−1  N p−1 1 p N  n=1 λ n n  m=1 c n,m a m  m  k=1 c m,k a k  p−1 = N p−1 1 p N  m=1  m  k=1 c m,k a k  p−1 a m N  n=m λ n c n,m . 122 Chapter 2. Inequalities Related to Hardy’s Inequality Hence, using Hölder’s inequality, we have N  n=1 λ n  n  m=1 c n,m a m  p  N p−1 1 p  N  m=1 λ m  m  k=1 c m,k a k  p  1/q  N  m=1 λ 1−p m  N  n=m λ n c n,m  p a p m  1/p with q = p/(p − 1), which by standard computation gives assertion (a). (b) By Lemma 2.2.4 we have, for C =(c m,k ) ∈M 3 , N  m=1 λ m  N  n=m c n,m a n  p  p N  m=1 λ m N  n=m c n,m a n  N  v=n c v,m a v  p−1  N p−1 3 p N  m=1 λ m N  n=m c n,m a n  N  v=n c v,n a v  p−1 = N p−1 3 p N  n=1  N  v=n c v,n a v  p−1 a n n  m=1 c n,m λ m . Hence, using Hölder’s inequality, we have N  m=1 λ m  N  n=m c n,m a n  p  N p−1 3 p  N  n=1 λ n  N  v=n c v,n a v  p  1/q  N  n=1 λ 1−p n  n  m=1 c n,m λ m  p a p m  1/p , where q = p/(p − 1) which by standard computation gives assertion (b). (c) Using Lemma 2.2.3 with an index n for which a N > 0, we obtain N  n=1 λ n  N  v=n c n,v a v  p  p N  n=1 λ n N  v=n c n,v a v  N  k=v c n,k a k  p−1  N 1−p 2 p N  n=1 λ n N  v=n c n,v a v  N  k=v c v,k a k  p−1 [...]... where b0 = a0 Lastly, from (an an ) = an 2 an + an 2 an + ( an )2 , we obtain G + B + F = −a0 b0 Hence 2 − 1 (a0 + B) 2 A D= + B) 2 − 1 (a0 2 G = ABC + B 2 − 1 (b0 2 G 2 − 1 (b0 2 + C) + C) C 1 2 1 2 2 a + B b0 + C G − A b 0 + C 2 0 4 2 1 2 2 − BG2 − C a0 + B 4 Therefore 2 2 2ABC + a0 + B b0 + C G − 2BG2 1 2 A b0 + C 2 2 1 2 + C a0 + B 2 2 2 a0 + B b0 + C 2 (AC), by the arithmetic mean–geometric... Inequality Therefore 2B (AC) 2 2 a0 + B b0 + C − 2B 1 2 1 b0 − a0 b0 + C + 2B 2 2 2 2 2 2 = a0 b0 + 2a0 b0 B + 2B 2 + a0 C 2 = (a0 b0 + B )2 + B 2 + a0 C This result gives 2B (AC) − B 2 2 (a0 b0 + B )2 + a0 C Therefore, if G2 < AC, 2 (AC), B the required inequality The relation G2 = AC, that is, ∞ 0 2 an ∞ ∞ 2 an = 2 an 0 2 an 2 0 holds if and only if there exists a constant λ such that 2 an = λan If λ... at the last step occurs 2 2 only when A(b0 + C )2 = C(a0 + B )2 Hence 2ABC − 2BG2 2 By √ Cauchy’s inequality, G (AC) − G, to obtain 2B (AC) − G 2 2 a0 + B b0 + C AC If G2 < AC, we can divide through by (AC) + G 2 2 a0 + B b 0 + C Therefore 2B (AC) 2 2 a0 + B b0 + C − 2BG 2 2 = a0 + B b0 + C − 2B(B + G) + 2B 2 But 1 2 1 B + G = −a0 b0 − F = −a0 b0 + b0 + C 2 2 1 32 Chapter 2 Inequalities Related to... m=1 N n=1 n=1 1 /2 N N n=1 n=1 1 /2 N = m=1 n 1 /2 2 am 1 /2 m=1 N 2 am n=1 m=1 1 /2 N n=m 1 /2 2 (N − m + 1)am n=1 m=1 N N 2 nan + n=1 N +1 2 1 N 2 nan 1 2 1 /2 N 2 nan = m=1 n n=1 N = 2 am m=1 N 1 /2 n 1 2 nan n=1 = am 1 n 2 nan 2 1 /2 n 1 n 2 nan 2 (N − n + 1)an n=1 N 2 an n=1 This proves the required inequality in (2. 3. 42) By taking zm = am and α = p + q in the following inequality (see [22 6]) α n zm m=1... b(x, y) 2 (m, n) x=1 y=1 M p p (m − 1) 2 (m − 1, n) − m 2 (m, n) m=1 1 p M 2 (M, n) p−1 0 (2. 3 .22 ) 138 Chapter 2 Inequalities Related to Hardy’s Inequality From (2. 3 .22 ) and by following the same procedure below (2. 3.16) up to (2. 3.18), we get M p M p p−1 p 2 (m, n) m=1 1 m m=1 m p n b(x, y) (2. 3 .23 ) x=1 y=1 From (2. 3.19) and (2. 3 .23 ), we observe that 2p M p p−1 SML L m−p p α3 (m, n), m=1 (2. 3 .24 ) n=1... OROLLARY 2. 2 .2 (Copson’s inequality, Theorem 2. 2.3) If p > 1 > c 0, λn and Λn are as in Theorem 2. 2.7, xn 0 and ∞ λn xn is convergent, then n=1 ∞ m=1 λm Λ−c m p ∞ λn xn n=m 1/p p 1−c ∞ 1/p p−c p λm Λ m xm m=1 P ROOF This proof follows from Corollary 2. 2.1, but with t 1 and 0 < m in (2. 2.34) and Theorem 2. 2.7 used instead of Theorem 2. 2.6 n 128 Chapter 2 Inequalities Related to Hardy’s Inequality 2. 3 Series... p−1 SML M n−p p (2. 3.19) b(x, y) n=1 2 (m, n), (2. 3 .20 ) s=1 where 2 (m, n) = 1 m m s=1 1 s s n x=1 y=1 From (2. 3 .20 ) and using inequality (2. 3.4), it is easy to observe that p 2 (m, n) − m 1 p p−1 m 1 p−1 n p−1 b(x, y) 2 (m, n) x=1 y=1 p p (m − 1) 2 (m − 1, n) − m 2 (m, n) (2. 3 .21 ) Keeping n fixed in (2. 3 .21 ) and letting m = 1, , M, and adding the inequalities, we have M p p−1 p 2 (m, n) − m=1... m b(x, y) (2. 3 .25 ) y=1 x=1 From (2. 3 .25 ) and using (2. 3.4), we observe that m p p−1 p α3 (m, n) − 1 p−1 p−1 b(x, n) α3 (m, n) x=1 p p (n − 1)α3 (m, n − 1) − nα3 (m, n) (2. 3 .26 ) Now, by following the same procedure below (2. 3.15) up to (2. 3.18), we get L p α3 (m, n) n=1 p L p p−1 p m b(x, y) (2. 3 .27 ) n=1 x=1 From (2. 3 .24 ) and (2. 3 .27 ), we observe that SML p p−1 3p L M p α4 (m, n), (2. 3 .28 ) n=1 m=1... say) (2. 3.37) 1+p/q (2. 3.38) and r r cn − cn+1 r−1 rcn an where r ( 1) is to be chosen later Letting θ = (2p − 1)−1 so that 0 < θ be rewritten as , 1, the left-hand side of (2. 3.36) may p(1−θ) pθ q bn cn L= bn n Applying Hölder’s inequality with indices 2/ (1 − θ ) and 2/ (1 + θ ), we have 2p (1−θ) /2 2pθ/(1+θ) 2q/(1+θ) cn bn L n n 2p = (1+θ) /2 bn (1−θ) /2 (1+θ) /2 r bn cn bn n , n where we have set (2p −... n) x=1 From (2. 3 .29 ) and using inequality (2. 3.4), we observe that p α4 (m, n) − p p−1 b(m, n)α4 (m, n) p−1 (2. 3 .29 ) 2. 3 Series Inequalities Related to Those of Hardy, Copson and Littlewood 1 p−1 p p (m − 1)α4 (m − 1, n) − mα4 (m, n) 139 (2. 3.30) Now, following the same procedure below (2. 3 .21 ) up to (2. 3 .23 ), we get M p M p p−1 p α4 (m, n) m=1 b(m, n) (2. 3.31) m=1 From (2. 3 .28 ) and (2. 3.31), we observe . + 1 2  a 2 0 +B  b 2 0 +C  G − 1 4 A  b 2 0 +C  2 −BG 2 − 1 4 C  a 2 0 +B  2 . Therefore 2ABC +  a 2 0 +B  b 2 0 +C  G −2BG 2  1 2 A  b 2 0 +C  2 + 1 2 C  a 2 0 +B  2   a 2 0 +B  b 2 0 +C   (AC), by.   a 2 0 +B  b 2 0 +C  −2B  1 2 b 2 0 −a 0 b 0 + 1 2 C  +2B 2 = a 2 0 b 2 0 +2a 0 b 0 B +2B 2 +a 2 0 C = (a 0 b 0 +B) 2 +B 2 +a 2 0 C. This result gives 2B  (AC) −B 2  (a 0 b 0 +B) 2 +a 2 0 C. Therefore, if G 2 < AC, B  2  (AC), the.   a 2 0 +B  b 2 0 +C  −2BG =  a 2 0 +B  b 2 0 +C  −2B(B +G) +2B 2 . But B +G =−a 0 b 0 −F =−a 0 b 0 + 1 2 b 2 0 + 1 2 C. 1 32 Chapter 2. Inequalities Related to Hardy’s Inequality Therefore 2B  (AC)   a 2 0 +B  b 2 0 +C  −2B  1 2 b 2 0 −a 0 b 0 + 1 2 C  +2B 2 =